Macedonian Junior Mathematical Olympiad

Let $y_k=x_k{x}_{k+1}{x}_{k+2}{x}_{k+3}$ (clearly $y_{n-2}=x_{n-2}{x}_{n-1}{x}_n{x}_1$, $y_{n-1}=x_{n-1}{x}_n{x}_1{x}_2$, $y_n=x_n{x}_1{x}_2{x}_3$). All $y_k$ are $1$ or $-1$. Then, by the conditions of the problem, we get $y_1+...+y_n=0$. Therefore $n=2k$ and moreover exactly $k$ of the numbers $y_1,...,y_n$ are equal to $1$ and the remaining $k$ are equal to $-1$. But then $y_1\cdot ...\cdot y_n=(-1{)}^k$. On the other hand, notice that $y_1\cdot ...\cdot y_n=x_1^4\cdot x_2^4\cdot ...\cdot x_n^4$, from where we get that $y_1\cdot ...\cdot y_n=1$, hence $k=2t$, i.e. $n=4t$.