G3.1 Let $f$ be a function such that $f(1)=1$ and for any integers $m$ and $n, f(m+n)=f(m)+f(n)+m n$. If $a=\frac{f(2003)}{6}$, find the value of $a$.
$\begin{array}{l}f(n+1)=f(n)+n+1=f(n-1)+n+n+1=f(n-2)+n-1+n+n+1=\ldots=1+2+\ldots+n+n+1 \\ \therefore f(n)=1+2+\ldots+n=\frac{n(n+1)}{2} \\ \frac{f(2003)}{6}=\frac{2004 \times 2003}{12}=334501\end{array}$