$2 \cdot 79$ Determine the smallest integer $n \geqslant 4$, from which any $n$ different integers, 4 different integers $a, b, c, d$ can be selected such that $a+b-c-d$ is divisible by 20.
(39th International Mathematical Olympiad Preliminary Question, 1998)
[Solution] Since $C_{7}^{2}=\frac{7 \times 6}{2}=21$,
it follows that from 7 different integers, one can definitely select a pair of integers that are congruent modulo 20. Let this pair of integers be $a, c$, then $a \equiv c(\bmod 20)$.

After selecting $a, c$ from the 7 different integers, there are still 5 different integers left. Adding 2 more integers that are different from the above 7 integers, we again form a set of 7 different integers. By the same analysis, we can select another pair of integers (let them be $b, d$), such that $b \equiv d(\bmod 20)$.
From (1) and (2), we get $a+b \equiv c+d(\bmod 20)$.
Thus,
$$
a+b-c-d \equiv 0 \quad(\bmod 20),
$$

Therefore,
$$
20 \mid a+b-c-d \text {. }
$$

Hence, from any 9 different integers, one can select 4 different numbers $a, b, c, d$ such that $20 \mid a+b-c-d$.
On the other hand, we can find 8 numbers, for example,
$$
0,1,2,4,7,12,20,40
$$

In these 8 numbers, no 4 numbers $a, b, c, d$ can satisfy
$$
20 \mid a+b-c-d .
$$