Example 8 Let $S=\{1,2, \cdots, 1000\}, A$ be a 201-element subset of $S$. If the sum of the elements in $A$ is divisible by 5, then $A$ is called a good subset of $S$. Find the number of all good subsets of $S$.
Solve: Divide all 201-element subsets of $S$ into 5 categories: $S_{0}, S_{1}, S_{2}, S_{3}, S_{4}$, where $S_{k}=\{A \mid A \subseteq S$, $|A|=201$ and the sum of elements in $A$ is congruent to $k \mod 5 \mid(k=0,1,2,3,4)$. Define the mapping
$f: S_{0} \rightarrow S_{k}(1 \leqslant k \leqslant 4)$ as follows: If $A=\left\{a_{1}, a_{2}, \cdots, a_{201}\right\} \in S_{0}$, then $f(A)=\left\{a_{1}+k, a_{2}+k, \cdots, a_{201}+k\right\}$, where if $a_{i}+k>1000$, replace $a_{i}+k$ with $a_{i}+k-1000$. It is not hard to see that $\sum_{i=1}^{201}\left(a_{i}+k\right)=\sum_{i=1}^{201} a_{i}+201 k \equiv k(\bmod 5)$, i.e., $f(A) \in S_{k}$. It is easy to verify that $f$ is a bijection, so $\left|S_{k}\right|=\left|S_{0}\right|(1 \leqslant k \leqslant 4)$. Let this number be $a$, then
$$
5 a=\left|S_{0}\right|+\left|S_{1}\right|+\left|S_{2}\right|+\left|S_{3}\right|+\left|S_{4}\right|=C_{1000}^{201} \text {. }
$$

Therefore, the number of good subsets of $S$ is $\left|S_{0}\right|=a=\frac{1}{5} C_{1000}^{201}$.