Peter wrote: Determine the smallest integer $ n \ge 4$ for which one can choose four different numbers $ a, b, c,$ and $ d$ from any $ n$ distinct integers such that $ a + b - c - d$ is divisible by $ 20$ . For n = 8, we have for example {0, 20, 40, 1, 2, 4, 7, 12} which has no two pairs with the same sum mod 20. Now suppose n > 8. If there are 7 distinct residues mod 20, then these form 21 pairs, so there must be two pairs with the same sum mod 20. So there can only be 6 or less distinct residues. Hence either there are 4 numbers a = b = c = d mod 20, or there are two pairs of numbers a = c mod 20 and b = d mod 20. In either case, we have $ a + b \equiv c + d$(mod 20)

For any $n\ge 9$, 4 nos. can be chosen from an set of $n$ number such that the assertion is true. My proof involves, $\binom {9}{2}$ and $\binom{7}{2}$ both being greater than 20, PHP asserts the required statement. From $\binom {9}{2}>20$, we havethat sum two pairs of numbers have the same residue modulo 20. If not then two integers exist such that $a\equiv b(mod20)$ Now remove $(a,b)$ and apply the same logic. The conclusion follows. Also, since $\binom {6}{2}=15<20$, this confirmation cannot be true for all sets of 8 numbers.