Answer is no A construction for counter example can be as follow : Let $g$ be a primitive root of $2003$ (fortunately it's prime) and set $a_i$ to be $g^i$ $\mod 2003$ . Then if there is such pair for $n > m$ we have : ${\frac{g^n + g^m}{2} \equiv g^{\frac{n+m}{2}} \Leftrightarrow \frac{g^{n-m} + 1}{2} \equiv g^{\frac{n-m}{2}} \Leftrightarrow (g^{\frac{n-m}{2}}} -1)^2 \equiv 0$ But $g$ is a primitive root , so above relation hence $\frac{n-m}{2}=2002k$ and as $0<m<n<2003$ this force $m$ to be equal to $n$ , contradiction .