.
Consider these integers moduo 4.
If they are all different modulo 4, then they are
p= 0 (mod4), q= 1 (mod4), r= 2(mod4) and s= 3(mod4).
In this case, we can choose two pairs from 6 differences, that produce 2 (mod4): r-p and s-q.
If, in opposite, they are not all different (mod4), then we have at least one pair producing difference 0(mod4).
So, in any case, we have at least one multiplier 0 (mod4).
Similar logic works for (mod3).
Of the four remainders mod3, we have at least one pair with the difference 0 (mod3).
Thus, the statement is proved.