Example 5 Find the number of integers $n$ greater than 1, such that for any integer $a$, $n \mid a^{25}-a$.

Find the number of integers $n$ greater than 1, such that for any integer $a$, $n \mid a^{25}-a$.
Analyzing Example 4, conjecture the maximum value of $n$ is 2730.
Solution: Let the set of positive integers that satisfy the condition be $S$. If $n \mid a^{25} - a$ and $m \mid a^{25} - a$, then $[m, n] \mid a^{25} - a$. Therefore, the least common multiple of all numbers in $S$ also belongs to $S$, meaning the largest number in $S$ is a multiple of every other number in $S$. If $m \mid a^{25} - a$, then all divisors of $m$ also divide $a^{25} - a$. Thus, it is only necessary to determine the largest number $m$, whose divisors greater than 1 form the set $S$.

Since $m \mid 2^{25} - 2$ and $m \mid 3^{25} - 3$, and $\left(2^{25} - 2, 3^{25} - 3\right) = 2 \times 3 \times 5 \times 7 \times 13$, by Example 4, using Fermat's Little Theorem, it is easy to prove that $2 \times 3 \times 5 \times 7 \times 13 \mid a^{25} - a$. Therefore, $m = 2 \times 3 \times 5 \times 7 \times 13$, and the set $S$ has a total of 31 elements.