TITLE: Combinatorics proof (any set of 16 numbers from 1 to 100 contains repeated sums)
QUESTION [10 upvotes]: Suppose that $A$ is a set of 16 distinct natural numbers and that $1\leq p\leq100$ for every $p$ in $A $.
Prove that $A$ contains 4 different numbers $a$, $b$, $c$, and $d$, such that $a+b=c+d$.

REPLY [2 votes]: a computer search shows that there is a set of 13 numbers between 1 and 100 with no four satisfying a+b=c+d: $$(1, 2, 12, 18, 22, 35, 43, 58, 61, 73, 80, 85, 87)$$ (and 87 is the lowest maximum that is achievable), but that the lowest possible maximum for a set of 14 positive integers with no a+b=c+d is 106: $$(1, 2, 7, 15, 28, 45, 55, 67, 70, 86, 95, 102, 104, 106)$$ so in fact, any set of 14 integers in the range 1 to 100 has repeated sums.
(I would have added this as a comment, rather than an answer, but I am new to math.stackexchange, so haven't accumulated the requisite reputation to comment)