Romanian Mathematical Olympiad

If $n$ is odd, then $-1$, $1$, $1$ are three complex roots of order $n$ of the unity, adding up to $1$.

On the other hand, if $x$, $y$, $z \in \mathbb{C}$, $x^n = y^n = z^n = 1$ and $x + y + z = 1$, then $|x| = |y| = |z|$, hence $\overline{x} + \overline{y} + \overline{z} = 1/x + 1/y + 1/z = 1$, which leads to $xy + xz + yz = xyz$.

Replacing $z = 1 - x - y$ gives $(x + y)(1 - x)(1 - y) = 0$, whence one of the numbers $x$, $y$, $z$ is $1$ and the other two are opposite.

In the case $n$ = odd, there are no opposite roots, so the final answer is: $n$ = even.