IMO 1994- Problem 1

Problem: Let $m$ and $n$ be positive integers. Let $a_1,a_2,...,a_m$ be distinct elements of $\left\{1,2,...,n\right\}$ such that whenever $a_i+a_j\le n$ for some $i,j,1\le i\le j\le m$, there exists $k$, $1\le k\le m,$ with $a_i+a_j=a_k$. Prove that: $\frac{a_1+a_2+...+a_m}{m}\ge \frac{n+1}{2}$.
Solution: Without loss of generality, we may assume that $a_1>a_2>...>a_m$. We claim that $a_i+a_{m+1-i}\ge n+1$ for $i=1,...,m$. The result then follows readily. To prove the claim ,we assume that on the contrary that it's false. Thus there exists $i$ such that $a_i+a_{m+1-i}<n+1$. Then $a_i<a_i+a_m<a_i+a_{m-1}<...<a_i+a_{m+1-i}\le n$. Thus $\left\{a_i+a_m,a_i+a_{m-1},...,a_{i}+a_{m+1-i}\right\}\subseteq \left\{a_1,a_2,...,a_{i-1}\right\}$
which is impossible. Thus the claim follows.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf