Hi. If you are just sending a signal over 100 feet of 22 or 24 guage wire simply to light LEDs, no sweat. You will of course need a cable with at least three conductors, two for the signals and one for return. Unless you want to get tricky and connect the LEDs up in opposite polarities, in which case you can use just two wires to select one LED or the other, but not both, by putting opposite signal levels across the wires.
At most the resistance across decent wire at that distance is 3 to 5 Ohms. Heavier guage will be less.If you need to assure yourself of this, simply measure the wire's resistance before you install it. You can expose and twist together the two wires at one end, then measure the resistance across the these wires at the other end. Face it, you're going to put 330 to 1000 Ohm resistors in line with the LEDs anyways for current limiting. If you have 2 LEDs and each pulls 20mA max, you're pulling 40mA max. The Vdrop across 200 feet of wire round trip, at say 5 Ohms, would be 0.1 Volt. This leaves you 4.9 Volts for LEDs which usually only need 2 to 2.3 Volts to work. Just wire the whole circuit ahead of time and take the measurements for Vdrop and current draw, then you'll know for sure.
If you plan to do something else then let us know NOW.
Later!
kenjj