If it's going to be near a swimming pool, definitely keep the voltage as low as possible, for safety. There's NO WAY I would put 60V anywhere near a swimming pool; even 15V could be dangerous. I think Steve's suggestion would be best - a two-wire bus carrying around 3-4V, with an individual current limiting resistor for each LED. It's not ideal from the point of view of efficiency, but safety is the most important concern.
So you would make up individual modules containing an LED and a limiting resistor, and run a two-wire bus around the pool, with the modules connected across the bus at various points along the length of the bus.
This also means you can add and remove LEDs fairly easily.
I'd suggest around 6.8 ohms for the limiting resistors. That means that at 50 mA LED current, they will drop 0.34 volts, which is many times the forward voltage variation from one LED to another, and this would make the brightnesses very even.
To get the LEDs to light to a good brightness, you'll need at least 20 mA per LED, or 400 mA total, which is quite a lot for AA batteries. I would use three rechargeable D cells in series, which gives 3.6V. That might be too bright; if so, you can add a switchable low-resistance voltage dropper in series with the battery, using rotary switch with several low-value resistors. This will let you select the brightness you want, and by dropping the current it will slightly lengthen the amount of time before the batteries need to be recharged.
I would remove the batteries from the circuit and recharge them separately. You don't want anything that's connected to mains power to be connected to your LED loop at any time.
Also your loop is quite long, so you should consider the resistance of the wire in the loop. If the wires are thin, you may get significant voltage drop across the length of the bus, which will affect the evenness of the LED brightness. Remember there will be voltage lost across EACH wire. If you connect your voltage to BOTH ends of the loop (i.e. make it an actual loop), you will greatly reduce the voltage loss.
For a bad case example, for a twin wire 100 feet long (not a loop) with 100 mA loading at the far end, if your wire has a resistance of 0.01 ohms per foot the total resistance to the end will be 0.01 ohms x 100 feet x 2 conductors = 2.0 ohms, so the voltage loss will be V = 2.0 ohms (resistance) x 0.1 (current) = 0.2V which would be noticeable. So don't use very thin wires for your loop, and connect it as a loop, i.e. connect both ends of the long wire together.
You might also want to add a fuse (e.g. 2A) in your battery supply in case the bus gets shorted out by being squashed or trodden on. And consider the possibility of people tripping on the wire, or hurting their feet by standing on the LEDs. This is especially important because there are many danger factors: pools are often used at night where there's poor visibility, there are hard edges and an immediate drowning danger, there may be alcohol involved, people are relaxed and may not be fully aware of their envornment, and so on.
Good luck and always think of safety!