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 = \dots = x_{n - 1} = 0 $ and $ x_n = -1 $. Then $ x_1 + x_2 + \dots + x_n = 0 $ and \[x_1 x_2 + x_2 x_3 + x_3 x_4 + \dots + x_{n - 1} x_n + x_n x_1 = 1.\]Thus, $ n = 3 $ and $ n = 4 $ are the only values that work, giving us $ \boxed{2} $ possible values of $ n $.