In parallel means they're all connected "across each other". For example, put them all side-by-side and twist all the wires on one side together, then twist all the wires on the other side together.
When you put several resistors of the same value in parallel, the resulting resistance is the resistance of one resistor divided by the number of resistors.
So four 2k7 resistors in parallel would be 2700 / 4 ohms, which is 675 ohms, which is outside the range for 620 ohms +/- 5%. So no, you can't.
With that value you could make up series-parallel arrangements to get closer to the value you need. To do this, you need the extended version of the parallel resistor formula, which is:
1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ...
Start with four 2k7s in parallel, that's 675 ohms. To reduce that to 620 ohms you use that formula:
1/620 = 1/675 + 1/needed_resistor.
This rearranges to needed_resistor = 1/ (1/620 - 1/675)
which is 7609 ohms. Three 2k7s in series will get you 8k1 (resistors in series just add). Substituting 8100 into that formula gives:
1/Rtotal = 1/675 + 1/8100
Rtotal = 623 ohms - good.
This arrangement needs seven 2700 ohm resistors. Four of them are directly in parallel with each other, and the other three are connected in series, and the series string is connected in parallel with the other four resistors as well.
There is a shorthand for writing resistors in parallel, using double vertical line symbols. The formula for that resistor combination is:
2700 || 2700 || 2700 || 2700 || (2700 + 2700 + 2700)
Now you can double-check it using the full formula for parallel resistors:
1/Rtotal = 1/2700 + 1/2700 + 1/2700 + 1/2700 + 1/(2700+2700+2700)
Rtotal = 623 ohms.
I hope this makes some sense. It is an option, although it will look pretty messy.