USAMO 2009

Problem: Let $n$ be a positive integer. Determine the size of the largest subsets of $\left\{-n,-n+1,...,n-1,n\right\}$ which does not contain three elements $a,b,c$( not necessarily distinct) satisfying $a+b+c=0$.
Solution: We claim the answer is $n$ for $n$ even and $n+1$ for $n$ odd. In both cases, we simply take the subsets of all odd numbers as an equality case.
Let $P$ be the set of positive numbers in our subset, and $N$ be the set of negative numbers( note that $0$ is not in our subset since $0+0+0=0$). Now, the condition is equivalent to saying that $-P,-N,P+N$ are disjoint.
Note that: $P+N=\left\{p+n:p\in P,n\in N\right\}$.
We now claim that $|P+N|\ge |P|+|N|-1$. To do this, let $P=\left\{x_1<...<x_m\right\}$ and let $N=\left\{y_1<...<y_n\right\}$. Then, the following elements are in $P+N:$
$x_1+y_1<x_2+y_1<...<x_m+y_1<x_m+y_2<...<x_m+y_n$
This proves the claim. Now, since $-P,-N,P+N$ are disjoint, we have that:
$|-P|+|-N|+|P+N|\le 2n+1\implies |P|+|N|+|P|+|N|-1\le 2n+1$
$\implies |P|+|N|\le n+1$.
This completes the proof for the odd case, but this is not strong enough for the even case.
We will now show that in fact $|P|+|N|\le n$ if $n$ is even. Assume not, so then equality must hold in all the above inequalities. Therefore, $(-P)\cup (-N)\cup (P+N)=\left\{-n,...\right\},$ and $|P+N|=|P|+|N|-1$. To start, since $0\notin P,N$, we must have that $n,-n\notin P+N$, so $n\in P$ and $-n\in N$.
Say that some $x\in P$. Then, since $-n\in N$, we must have that $x-n\in P+N$. However, since $-P,-N,P+N$ are disjoint, we must have that $x-n\notin -P,$ or $n-x\in P$. Thus,
$x\in P\implies n-x\notin P$. Therefore, one seems that from the numbers $1,...,n$, at most $1+\frac{n-2}{2}=\frac{n}{2}$ of them can be in $P$(note that $\frac{n}{2}\notin P$ since $\frac{n}{2}\in P\implies n-\frac{n}{2}\notin P$). Similar, one can show that there are at most $\frac{n}{2}$ elements in $N$. However, this implies $|P|+|N|\le n,$ which is a contradiction, since we started by assuming $|P|+|N|>n$. Therefore, for $n$ even, we can have at most $n$ elements in the subset.
Source: https://artofproblemsolving.com/community/c473124h1546246_more_combinatorics