XXIV OBM

Let $p$ be a prime and $A=\{p,2p,3p,\dots ,2002p\}$. The sum of any quantity of numbers from $A$ is at most $p+2p+\cdots +2002p=1001\cdot 2003p$. Choose any $p>1001\cdot 2003$ and we are done, because every sum of numbers from $A$ is a multiple of $p$ but not of $p^2$, and cannot be a perfect power.