Right now I am reading about comprators.
Nah, c'mon, ya' can do this with one transistor. I think.
4x3.5 = 14v needed by the LEDs.
There is 1v drop in the emitter follower,
so if the LEDs want 30 mA then R2 = [(18-1)-14/.03] = 100Ω
Figure base current is 30/10 mA or less.
R1 = 0 ohms.
R4 = inf.
The LDR threshold is sqrt (15x500) = 87k at which point the LEDs must just see 14v.
14v + Vbe = ~15v = base voltage.
15/87k = only 0.2 mA in the base divider resistor network, and this is not the recommended 5x the base current, so you may need a Darlington for the transistor. Try it anyway, it may work.
Anyway, that leaves 3v across R3. 3v/0.2 = 15k =R3.
BTW, sqrt (15x500) is the geometric, not the arithmetic, average. I use it for widely varying values like this.
I can't guarantee what LED current you will get with only a 100Ω series resistor, but there is so much slop in this circuit it may work anyway.
Another way to look at this is that the circuit supplies no voltage gain and the output load needs 30 mA.
The input can supply 18(87k)/(87k +15k) = 15v with a source impedance of 87k||15k = 13k so the source can supply an input current of 15/13k = 1.2 mA.
The current gain of the circuit @ 30mA needs to be 30/1.2 = 25. One transistor might do this after all.
With 500k max resistance and 14v this is a current through the LDR of 28 uA. (18-14)/28 uA = 140k.
From your description the max value of R3 may need to be >140k.