Example 16 (1991 Australian Mathematics Communication Competition Question) Find the smallest positive integer $n$ such that for any selection of $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
Consider the set of 996 integers $M=\left\{a_{i} \mid a_{i}=0,1,2, \cdots, 995\right\}$, then for all $i \neq j, i, j=0,1,2, \cdots, 995$,
$$
a_{i}+a_{j} \leqslant 995+994=1989 \text {, }
$$
$0996$, so there must exist at least two different numbers $a_{i}, a_{j}$, such that $a_{i}=-a_{j}$, thus we have
$$
1991 \mid\left(a_{i}+a_{j}\right) .
$$