The set S consists of 20 distinct positive integers, each below 70.
Create a set S' of 190 pairwise absolute values of differences between two distinct members of S.
Prove that in S' there must exist at least one value that appears more than 3 times.
Forget the 190 couples. Order the elements of S in ascending order and consider just the 19 pairwise differences between adjacent elements. Among just these 19 there must be one number that appears at least 4 times!!
PROOF: Assume that no number appears 4 or more times. Then the minimum value of the sum of the 19 differences between adjacent elements must be 3*1 + 3*2 + 3*3 + 3*4 + 3*5 + 3*6 + 7 = 3*21 + 7 = 70. But this sum is just the difference between the highest and lowest number in S, and that can be at most 69. So we have a contradiction, and our initial assumption is wrong. Thus, at least one number appears at least 4 times among the 19 pairwise differences. q.e.d
(ASIDE: I expect that there are many more than just one number that must appear more than 3 times among the full 190 elements, but I am moving on.)
A much stronger proof (spoiler)
