jmc

We claim that the only such positive integers $n$ are 3 and 4.

For $n=3,$ $x_1+x_2+x_3=0.$ Then $(x_1+x_2+x_3{)}^2=0,$ which expands as $x_1^2+x_2^2+x_3^2+2(x_1{x}_2+x_1{x}_3+x_2{x}_3)=0.$ Therefore, $$x_1{x}_2+x_2{x}_3+x_3{x}_1=-\frac{1}{2}(x_1^2+x_2^2+x_3^2)\le 0.$$For $n=4,$ $x_1+x_2+x_3+x_4=0.$ Then $$x_1{x}_2+x_2{x}_3+x_3{x}_4+x_4{x}_1=(x_1+x_3)(x_2+x_4)=-(x_1+x_3{)}^2\le 0.$$For $n\ge 5,$ take $x_1=-1,$ $x_2=0,$ $x_3=2,$ $x_4=x_5=\cdots =x_{n-1}=0$ and $x_n=-1.$ Then $x_1+x_2+\cdots +x_n=0$ and $$x_1{x}_2+x_2{x}_3+x_3{x}_4+\cdots +x_{n-1}{x}_n+x_n{x}_1=1.$$Thus, $n=3$ and $n=4$ are the only values that work, giving us $\boxed{2}$ possible values of $n.$