Let $p$ be a prime and $A = \{p, 2p, 3p, \dots, 2002p\}$. The sum of any quantity of numbers from $A$ is at most $p + 2p + \dots + 2002p = 1001 \cdot 2003p$. 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.