smc

The first $2003$ even counting numbers are $2,4,6,...,4006$. The first $2003$ odd counting numbers are $1,3,5,...,4005$. Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$. $(2+4+6+...+4006)-(1+3+5+...+4005)=(2-1)+(4-3)+(6-5)+...+(4006-4005)$ $=1+1+1+...+1=\boxed{2003}$