This might be dumb of me not to understand... or it might not be... umm... anyway, is it 2^(N-1) or (2^N) - 1?

If it's 2^(N-1), then the solution is all powers of two.

For 2^(N-1) to be divisible by N, N must have a prime factorization of only 2s (as 2^anything has only 2s).

It is easy to see that

2^((2^0)-1)/2^0 = 1

2^((2^1)-1)/2^1 = 1

2^((2^2)-1)/2^2 = 2

2^((2^3)-1)/2^3 = 16

and so on... but this seems pretty easy, which leads me to suspect that it's (2^N) - 1

Because (2^N) - 1 is odd, all even numbers can be ruled out.