4. Find the smallest positive integer $n$ such that for any $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
(1991, Australian Mathematical Olympiad)
Let $M=\left\{a_{i} \mid a_{i}=0,1, \cdots, 995\right\}$.
Since $a_{i}+a_{j} \leqslant 995+994=1989<1991$, $0<\left|a_{i}-a_{j}\right| \leqslant 995$,
then the sum and difference of any two numbers in $M$ are not multiples of 1991.
Therefore, $n \geqslant 997$.
Let $a_{1}, a_{2}, \cdots, a_{997}$ be any 997 integers.
If there exist $a_{i} 、 a_{j}$ such that $a_{i} \equiv a_{j}(\bmod 1991)$, then the difference $a_{i}-a_{j}$ is divisible by 1991.

If any two numbers $a_{i} 、 a_{j}$ are not congruent modulo 1991, consider the smallest absolute residues of 1991
$$
-995,-994, \cdots,-1,0,1, \cdots, 994,995 \text {. }
$$

Divide them into 996 groups:
$$
\{-995,995\},\{-994,994\}, \cdots,\{-1,1\},\{0\} \text {. }
$$

Thus, there are at least two different numbers $a_{i} 、 a_{j}$ such that
$$
a_{i}=-a_{j}(\bmod 1991) \text {. }
$$

In this case, 1991 ! $\left(a_{i}+a_{j}\right)$.