smc

We already know that $a_1=2001$, $a_2=2002$, $a_3=2003$, and $a_4=2000$. Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5=2002+2003-2000=2005$, $a_6=2003+2000-2005=1998$, $a_7=2000+2005-1998=2007$, and so on. We can now discover the following pattern: $a_{2k+1}=2001+2k$ and $a_{2k}=2004-2k$. This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002}=2004-2\cdot 1002=\boxed{0}$.