, 2015

Call *vertical* the $n$ edges that join corresponding vertices of the two bases. A vertical edge is *odd* if it has different number at its endpoints and *even* otherwise. Take two adjacent vertical edges $e_1$ and $e_2$. They determine a lateral face $F$ which is a rectangle with opposite sides $e_1$ and $e_2$. The product of the numbers at the vertices of $F$ is known to be $-1$. It is also the product of the numbers at the endpoints of $e_1$ multiplied by the respective product for $e_2$. Hence $e_1$ and $e_2$ are of different kinds, one is odd and the other is even. Thus odd and even vertical edges alternate, which is possible only if $n$ is even. Let $k_1$ and $k_2$ be the numbers of $-1$ at the vertices of the two bases. Both $k_1$ and $k_2$ are odd by hypothesis, hence the total number $k = k_1 + k_2$ of $-1$ is even. On the other hand $k$ has the same parity as the number of odd vertical edges, which by the above equals $n/2$. Hence $n/2$ is even, meaning that $n$ is divisible by $4$.

So such an assignment of $+1$ and $-1$'s is possible only if $n$ is a multiple of $4$. Conversely, let $n = 4k$, $k \geq 1$. Let the two bases $A_1A_2...A_{4k}$ and $B_1B_2...B_{4k}$. Assign a $-1$ to each of the vertices $A_1, A_2, ..., A_{4k-3}$ ($2k - 1$ of them), and also to $B_{4k-1}$. Write a $+1$ at all remaining vertices. The product of numbers on each face is $-1$. A vertical edge $A_iB_i$ is odd or even according as $i$ is odd or even, hence the product of numbers on each lateral face is also $-1$. So the assignment has the necessary properties. In conclusion, the answer to the question is: for all $n$ divisible by $4$.