To solve the problem, we are tasked with finding the number of integers $n > 1$ such that $n$ divides $x^{13} - x$ for every positive integer $x$.

First, observe that if $n \mid x^{13} - x$ for every integer $x$, then $n \mid x^{13} - x$ for each $x$ in particular values, such as $x = 0, 1, 2, \ldots, n-1$. This means that $n$ divides the polynomial $x^{13} - x$ evaluated at these integers.

An important observation is that the polynomial $x^{13} - x$ corresponds to the characteristic property of a finite field. Specifically, $x^{13} - x \equiv 0 \pmod{p}$ for a prime $p$ implies that $p \mid 13$ or the multiplicative order of $x \pmod{p}$ divides 13.

The roots of the polynomial $x^{13} - x \equiv 0 \pmod{n}$ are precisely the elements of the finite field $\mathbb{Z}_n$ if $n$ is a prime power.

The polynomial $x^{13} - x$ can be factored using:
$$x^{13} - x = x(x^{12} - 1) = x(x^6 - 1)(x^4 + x^2 + 1).$$

Notice that the polynomial $x(x^6 - 1)(x^4 + x^2 + 1)$ implies that $n$ should divide each of the factors, either directly or by induction that all prime divisors of $n$ must also be Fermat primes where necessary.

At this point, it is particularly significant that the prime divisors $n$ must satisfy $n \equiv 1 \pmod{13}$. Therefore, we need to find all integer divisors greater than 1 of order 13. This includes small prime powers such that for each prime $p$, $p \equiv 1 \pmod{13}$, which in the case of modulo 13 implies possibly restricted to to the factor set characteristics.

Ultimately, using the properties of congruences and finite fields, we find that:

For $n$ such that $n$ divides $x^{13} - x$ for all integers $x$, we have the specific minimal divisors governing congruence properties from derived direct or field characteristics:
$$n \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 28, 30, 32, 36, 40, 42, 48, 60, 63, 84\}$$

Hence, the number of such integers $n$ is:
$$\boxed{31}.$$