USAMO

Let $S$ be a subset of $\{-n,-n+1,\dots,n-1,n\}$ of largest size satisfying $a+b+c\neq 0$ for all $a,b,c\in S$ . First, observe that $0\notin S$ . Next note that $|S|\geq \lceil n/2\rceil$ , by observing that the set of all the odd numbers in $\{-n,-n+1,\dots,n-1,n\}$ works. To prove that $|S|\leq \lceil n/2\rceil$ , it suffices to only consider even $n$ , because the statement for $2k$ implies the statement for $2k-1$ as well. So from here forth, assume $n$ is even.
For any two sets $A$ and $B$ , denote by $A+B$ the set $\{a+b\mid a\in A,b\in B\}$ , and by $-A$ the set $\{-a\mid a\in A\}$ . Also, let $A_+$ denote $A\cap\{1,2,\dots\}$ and $A_-$ denote $A\cap\{-1,-2,\dots\}$ . First, we present a lemma:
Lemma 1 : Let $A$ and $B$ be two sets of integers. Then $|A+B|\geq|A|+|B|-1$ .
Proof : Write $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots,b_m\}$ where $a_1<\dots<a_n$ and $b_1<\dots<b_m$ . Then $a_1+b_1,a_1+b_2,\dots,a_1+b_m,a_2+b_m,\dots,a_n+b_m$ is a strictly increasing sequence of $n+m-1$ integers in $A+B$ .
Now, we consider two cases:
Case 1 : One of $n,-n$ is not in $S$ . Without loss of generality, suppose $-n\notin S$ . Let $U=\{-n+1,-n+2,\dots,n-2,n-1,n\}$ (a set with $2n$ elements), so that $S\subseteq U$ by our assumption. Now, the condition that $a+b+c\neq 0$ for all $a,b,c\in S$ implies that $-S\cap(S_++S_-)=\emptyset$ . Since any element of $S_++S_-$ has absolute value at most $n-1$ , we have $S_++S_-\subseteq \{-n+1,-n+2,\dots,n-2,n-1\}\subseteq U$ . It follows that $S_++S_-\subseteq U\setminus -S$ , so $|S_++S_-|\leq 2n-|S|$ . However, by Lemma 1, we also have $|S_++S_-|\geq |S_+|+|S_-|-1=|S|-1$ . Therefore, we must have $|S|-1\leq 2n-|S|$ , or $2|S|\leq 2n+1$ , or $|S|\leq n$ .
Case 2 : Both $n$ and $-n$ are in $S$ . Then $n/2$ and $-n/2$ are not in $S$ , and at most one of each of the pairs $\{1,n-1\},\{2,n-2\},\ldots,\{\frac n2-1,\frac n2+1\}$ and their negatives are in $S$ . This means $S$ contains at most $2+(n-2)=n$ elements.
Thus we have proved that $|S|\leq n=\lceil n/2\rceil$ for even $n$ , and we are done.