I suppose the constraint of $n$ is actually $2\leqslant n\leqslant 2^N$. The answer is $n=4,6,8,...,2^N$. For the prove, it's enough to show that there not exists odd-length cycle. This is obvious by noting that each edge connect two vertices whose coordinate differ by exactly one point. If there's odd cycle, then there's a coordinate that was changed by odd number of times through the cycle, which means the coordinate can't be back to the initial one. For the construction, simply use induction on $N$. Also use the fact that when $n=2^N-1$, we can start the cycle by any edge that is adjacent to vertex label $0$ (by virtue of symmetry of the hypercube.)