OK. I think R7 should be about 560 ohms, for example
http://www.digikey.com/product-detail/en/CFR-50JB-52-560R/560H-ND or
http://uk.farnell.com/welwyn/mfr4-560rfi/resistor-metal-film-560-ohm-500mw/dp/1833360 and R1 should be about 330 ohms 0.5W fusible - for example,
http://www.digikey.com/product-detail/en/FRM-50JR-52-330R/330DTCT-ND. Farnell don't have a 330 ohm fusible resistor but 470 ohms will do:
http://uk.farnell.com/te-connectivity-neohm/frn50j470r-s/resistor-fusible-470r-5-0-5w/dp/1898508
The most likely cause of R1's failure would be the failure of C4, and I strongly suggest that you replace C4, even if it measures OK. A suitable replacement is the EPCOS B32922C3224K which is available from Digikey:
http://www.digikey.com/product-detail/en/B32922C3224K289/495-4897-1-ND and Farnell:
http://uk.farnell.com/epcos/b32922c3224k189/capacitor-film-0-22uf-10-rad/dp/2367352
A better quality alternative for a slightly higher cost is the EPCOS B32912A3224K which is also available from Digikey:
http://www.digikey.com/product-detail/en/B32912A3224K/495-3970-ND and Farnell:
http://uk.farnell.com/epcos/b32912a3224k/capacitor-pp-film-0-22uf-330vac/dp/2098616
For safety, it is wise to connect a "bleed resistor" across C4, to discharge it when power is removed. There is no bleed resistor in this design but you can easily add one. I suggest adding two 100k 0.25~0.5W metal film resistors in series, on the underside of the board between the terminals of C4. Suitable resistors would be
http://www.digikey.com/product-detail/en/RNF12FTD100K/RNF12FTD100KCT-ND and
http://uk.farnell.com/multicomp/mcmf0w2ff1003a10/resistor-metal-film-100kohm-500mw/dp/1126841
The following image is a simulation of the important current paths in the circuit, using a relay coil resistance of 4608 ohms (a 500 mW coil), the value you measured on the original relay on your board.
The simulation shows that the average voltage across the relay coil is about 41V. This is pretty low for a 48V relay and I would recommend replacing the relay with one with a 360 mW coil (coil resistance 6400 ohms) such as a Panasonic JS1 series relay:
From Digikey (USA):
http://www.digikey.com/product-detail/en/JS1-F-48V-F/255-2889-ND/2177080
From Farnell (UK):
http://uk.farnell.com/panasonic-ew/js1-b-48v-ft/relay-pcb-spco-10a-48vdc/dp/2095679
The JS1 series has the same pin layout as the one in your board, but you'll need to check the dimensions and pin distances to be sure it will fit.
These are pretty cheap and using a relay with a more sensitive coil will improve safety. Variations in AC mains voltage will affect the relay coil voltage, and if the relay does not pull in firmly, the contacts can arc, causing damage and possibly fire. Using a 360 mW coil gives an average coil voltage of about 53V which is within the relay's specifications and safer than 41V.
Unfortunately, changing to a sensitive coil relay will increase the voltages in the circuit, and the 10 µF electrolytic, and the transistors, will be run uncomfortably close to their voltage limits. You should replace the 10 µF electrolytic with one rated for 100V, and the transistors with types rated for 100V and reasonably high gain, such as:
For the NPN: ZTX694B, available from Digikey:
http://www.digikey.com/product-detail/en/ZTX694B/ZTX694B-ND and Farnell:
http://uk.farnell.com/diodes-inc/ztx694b/transistor-npn-e-line/dp/9525610
For the PNP: ZTX795A, available from Digikey:
http://www.digikey.com/product-detail/en/ZTX795A/ZTX795A-ND or NTE129P, available from Farnell:
http://uk.farnell.com/nte-electronics/nte129p/bipolar-transistor-pnp-80v-to-237/dp/4524159
These transistors all have different pinouts from the originals.
The next image is the schematic redrawn to it's easier to follow and explain.
I've drawn the AC mains supply, the wall switch, and the fan on the right side of the diagram, as I think they should be connected. If I'm wrong about this, the following circuit description won't be exactly right.
The rail across the bottom, marked "0V", is the "zero volt" rail - the reference rail. It is not "ground", and you shouldn't use the ground symbol for it. It is connected to the Neutral connection from the AC mains, which is grounded at the switchboard, but it is not ground and should not use any kind of ground symbol.
Here is how I believe the circuit works.
With the circuit idle, and the wall switch OFF, there is 230V AC RMS (relative to the circuit's 0V rail) on CN2 but since RL1 is de-energised, it doesn't go anywhere. The circuit starts up when the wall switch is turned on.
When 230VAC appears on CN3 from the wall switch, C3 starts to charge up on every alternate half-cycle of the AC mains waveform. There are two paths for this charging current: through C4, R1 and D2 (D1 is also involved), and through R2 and D3.
C4, D1 and D2 form a "capacitor fed" power supply that charges C3 from the mains supply. R1 is included for safety; if C4 fails, R1 will overheat very quickly and go open-circuit, protecting the rest of the circuit from damage.
This capacitor fed power supply is a common way that small circuits can be powered from the mains. It avoids the size, weight, and cost of a transformer, but it provides no isolation, and is only suitable when the whole circuit can be kept isolated, and when its load current is small (this circuit draws only about 10 mA). The method is often used to power LEDs directly from the mains in space-sensitive and cost-sensitive applications.
Capacitor fed power supplies are described in many places on the web, but often not very well. Also, most designs use full-wave rectification, whereas this board uses half-wave rectification. Here is one explanation that isn't too bad:
http://www.dos4ever.com/TiT/TiT.html (search for Figure 24).
So current through C4 via D2 charges C3, but current through R2 and D3 also charges C3 while the switch is closed. D3 also clamps the peak voltage at point B in the circuit to slightly higher than the positive supply rail (marked V+).
D4 pulls point A up to roughly the V+ rail voltage as well, so C2 is discharged; there is no significant voltage across it. R9 and R12 form a voltage divider that sets point C to about 40% of the V+ voltage.
Remember, in this description, voltages at single points in the circuit are measured relative to the 0V rail at the bottom of the schematic.
Q2's base is therefore fed (via R11) from a higher voltage than its emitter, which causes Q2 to turn OFF (not conducting). R8 pulls Q1's base to 0V and Q1 turns OFF as well. Current flows through R7 and through the relay coil.
D7 is needed to protect Q2 against reverse base-emitter voltage, which can permanently damage the transistor if it exceeds about 7V. No significant reverse current can flow from R11 through D7.
When C3 has charged up to around 40V, there will be enough voltage across the relay coil for the relay to pull in. When this happens, its contact changes over from the way it's shown on the schematic to the normally open contact, which is fed from mains Phase directly. This feeds the mains to the fan and into C4 as before; the wall switch is now not part of the circuit.
But the wall switch continues to affect the voltage at point A due to current through R2 which produces a voltage that's clamped to the V+ rail and holds point A near the V+ rail voltage. This keeps Q2 and Q1 OFF, so the relay remains ON.
The circuit remains in this state while the wall switch is ON.
When the wall switch is turned OFF, the fan and the circuit remain powered (because the relay is active), but current can no longer flow through R2 and D4 into point A, so C2 starts charging up slowly, due to current flowing through R3 and VR1. As it charges up, the voltage at point A falls.
The voltage at point A falls at a rate determined by the capacitance of C2 and the total resistance of R3 plus VR1 at its present setting. Once this voltage has fallen below about 40% of the V+ voltage, as set by R9 and R12, Q2 now has forward bias, because its base is negative relative to its emitter, so it starts to conduct.
Conduction in Q2 quickly reaches the point where enough voltage appears across R8 for Q1 to start to turn ON. When this happens, the V+ rail voltage starts to fall, because significant current is being drawn from it via R7 and Q1. Since C2 has about 60% of the original V+ voltage across it, this dropping V+ voltage causes the voltage at point A to drop further, turning Q2 on harder and therefore saturating Q1.
This change happens quickly and Q1 pulls its collector down to 0V, removing almost all of the voltage at point B and across the relay coil, causing the relay to drop out. The relatively high current through R7 and Q1 discharges C3 fairly quickly, and since the relay has returned to the position shown in the schematic and the wall switch is OFF, there is no longer any mains voltage feeding into C4.
The V+ rail collapses fairly quickly. Q2 and Q1 remain ON until the rail voltage is so low that when Q1 turns OFF, there is not enough rail voltage available to activate the relay coil. The V+ rail then discharges all the way to zero through R9 and R12 and the circuit is idle once again, ready to be reactivated when the wall switch is turned ON.
Normally, relay coils have a diode connected across them to protect the driving device from "back EMF" when it turns OFF and the magnetic field in the relay coil collapses and produces a voltage spike because of its inductive nature. That diode isn't needed in this circuit because the current to the relay coil is interrupted by Q1 shorting the coil out. When Q1 finally turns OFF, there is almost no energy available in the V+ rail to have any effect.
The V+ voltage during operation is mainly determined by the relay coil's resistance in series with R7. The V+ voltage is not regulated or limited. The two current sources (C4 and R2) provide a reasonably accurately controlled amount of current onto the rail, and the resistance of R7 plus the relay coil determines how much voltage appears on V+ according to Ohm's Law.
I'm not sure what C5 does. It will help protect Q2's base-emitter junction from any spikes, and it may play a part in ensuring that Q2 and Q1 remain ON as the V+ rail is falling towards 0V. It's important that they remain ON during this time; if they turn OFF too quickly, there might still be enough voltage on the V+ rail to make the relay pull in.