1.18 Let $S=\{1,2, \cdots, 1990\}$, if the sum of all numbers in a 31-element subset of $S$ is a multiple of 5, then it is called a "good subset". Find the number of all good subsets of $S$.
For any subset $A$ of $S$, let $|A|$ and $\|A\|$ denote the number of elements in $A$ and the sum of all elements in $A$, respectively. Let
$$
F_{k}=\{A|A \subset S,| A \mid=31,\|A\| \equiv k(\bmod 5)\},
$$

where $k=0,1,2,3,4$.
For any $\left\{x_{1}, x_{2}, \cdots, x_{31}\right\}=A \in F_{0}$, define
$f_{k}(A)=\left\{x_{1}+k, x_{2}+k, \cdots, x_{31}+k\right\}, k=1,2,3,4$, where if $x_{i}+k>1990$, it is understood as $x_{i}+k-1990$. Since $31 k \equiv k, 1990 \equiv 0(\bmod 5)$, we have $f_{k}(A) \in F_{k}$. It is easy to see that the mapping $f_{k}$ is a bijection from $F_{0}$ to $F_{k}$, hence
$$
\left|F_{0}\right|=\left|F_{1}\right|=\left|F_{2}\right|=\left|F_{3}\right|=\left|F_{4}\right| \text {. }
$$

Since $F_{0}, F_{1}, F_{2}, F_{3}, F_{4}$ are pairwise disjoint and $\sum_{k=0}^{4}\left|F_{k}\right|=C_{1990}^{31}$. Therefore, the total number of good subsets of $S$ is $\left|F_{0}\right|=\frac{1}{5} C_{1990}^{31}$.