[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
$$20 \mid a+b-c-d .$$