Fact 1: Let $A\subseteq\{1,2,...,2n\}$. If $n+1\leq |A|$, then there exists 2 elements $a,b\in A$ such that $a+b=2n+1$.
Proof: This can be shown by writing $\{1,2,...,2n\}$ as the union of $n$ disjoint sets $\{i,2n+1-i\}$ where $1\leq i\leq n$. Since $n+1\leq|A|$, we know that $|A\cap \{i,2n+1-i\}|=2$ for some $i\in\{1,2,..,n\}$. Now we're done because $i+2n+1-i=2n+1$.
One also knows that the bound of fact 1 is sharp because $\{1,2,...,n\}$ does not have 2 elements which add up to $2n+1$.
Definition: Let $f(n)$ be the smallest number such that: $$\forall A\subseteq\{1,2,...,4n\}[|A|>f(n)\rightarrow \exists a,b,c,d\in A [a+b+c+d=2(4n+1)] ]$$
Question: What are good upper bounds on $f(n)$ ?
Fact 2: $f(n)\leq 3n+1$
Proof: We repeat the idea of the proof of fact 1. Let $A\subseteq\{1,2,...,4n\},|A|\geq 3n+1$. It can be shown that $\{1,2,...,4n\}$ can be written as the union of disjoint sets: $$\{1,2,...,4n\}=\cup_{i=1}^{n}\{i,2n+1-i,2n+i,4n+1-i\}$$ (It can also be shown that $\land_{i=1}^n[i,2n+1-i,2n+i,4n+1-i \,are\, distinct\,]$ ) Since $|A|\geq 3n+1$, we know that $|A\cap\{i,2n+1-i,2n+i,4n+1-i\}|=4$ for some $i\in\{1,2,...,n\}$. Now we are done because: $$i+2n+1-i+2n+i+4n+1-i=2(4n+1)$$ Thus $f(n)\leq 3n+1$.
Unlike fact 1, I am not sure if the bound of fact 2 is sharp. I think there might be more clever ways of using the pigeon hole principle to obtain better upper bounds on $f(n)$. Question: Are there better bounds for $f(n)$ other than those given by fact 2 ?
My main intrest is about seeing the more clever ways of using the pigeon -hole principle being applied to this problem and not the problem itself.
Thank you
$f(n)=2n+1$. We write $[1,4n]$ as the disjoint union of $2n$ pairs $\{i,4n+1-i\}$. If $|A|>f(n)$, then at least two pairs must have both of their elements in $A$; the sum of those four elements is $8n+2$. On the other hand, let $A=[1,2n+1]$. The four largest elements from $A$ sum to $8n-2$, so no four elements from $A$ sum to $8n+2$.
Edit: fixed typo.