We notice if $n$ is okay, then $n+2$ satisfies(add $x_{n+1}=1, x_{n+2}=-1$). Clearly, $2 \mid n$ satisfies. We can prove $n \neq 3$. Now we solve $n=5$. We select $x_1=x_2=x_3=-1$, $x_4=\frac{3+\sqrt{5}}{2}$, $x_5=\frac{3-\sqrt{5}}{2}$, then $x_1+x_2+...+x_5=\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_5}=0$. So $n \neq 3$ satisfies.