$x^{13} - x = x(x + 1)(x - 1)(x^2 - x + 1)(x^2 + x + 1)(x^2+1)(x^4-x^2+1)$. By Fermat's Little Theorem, this is divisible by $13$. Also, $x(x + 1)(x - 1)$ is divisible by $6$, and $x(x + 1)(x - 1)(x^2 + 1)$ is divisible by $5$. Finally, $x(x^6 - 1)$ is divisible by $7$ by Fermat's Little Theorem. Hence it is divisible by $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$. So there are $\boxed{31}$ integers (we must omit n=1).