Let $A=\{a_1,\dots,a_{2003}\}$ with $a_1<\cdots<a_{2003}$. Suppose the assertion is false, and that for $S\triangleq \textstyle \sum_i a_i$, $S$ is divisible by $a_i+a_j$ for $1\le i<j\le 2003$. We inspect the pairs $(a_i,a_{2003})$ for $1\le i\le 2002$. Note that since $a_i$ are distinct, it follows that $k_i\triangleq S/(a_i+a_{2003})$ are all distinct. Notice now that $a_1+a_{2003}<S<2003(a_{2002}+a_{2003})$, yielding that $k_i\in\{2,\dots,2002\}$, a range of $2001$ integers. SInce we have $2002$ $k_i$'s, two of them must be the same by pigeonhole, a contradiction.