For your problem, I would recommend the KISS principle (Keep It Simple). Stick 2 LEDs in series with a 200 ohm 1/2W resistor (preferably 1W) and run from the battery voltage through a switch and a fuse.
When battery voltage is 12.5V (fully charged, engine off), and the LEDs are hot, you are looking at 31mA. At the other extreme, when the battery is down to 11V and the LEDs are relatively cool, your current will be down to 22mA.
You can just run a power wire all the way out. At regular intervals, strip the insulation and solder a resistor. Then shrink wrap over the exposed part of the resistor & wire and move on to the next one. You will end up with a long wire with resistors sticking out at regular intervals. Attach two LEDs in series to the resistors. Take a similar ground wire, strip and solder the other end of the second LED, shrink wrap, and repeat the process. You will end up with a string of 50 pairs of LEDs in parallel. You are looking at about 1.2A current in the leg (25mA typical * 50 pairs), so you can get away with a 24 gauge wire, though I would prefer 22 gauge. The work sequence I have described will work out only if you solder/shrink-wrap in the proper sequence. So think this through before starting.
In this setup, you are wasting about 50% of the energy. If you were to use 3 LED strings and a control circuit for each string, you will bring the efficiency up to 66% and the power consumption down to 1.6A for the whole system versus 2.5A for this setup, but the whole thing will get much more expensive and complicated.
The reason I am suggesting 1W resistors is because if you have a problem and the non-heat-shrink side of the resistor gets shorted to the boat, you will see about 0.75W through the resistor. A 1W resistor will survive that. A 1/2W resistor will blow.
Good luck.
---55p
Edited to fix my stupid math mistake.