What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers? 
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$
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 $(2+4+6+...+4006)-(1+3+5+...+4005)$. 
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$
$= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$