Let the sequence be $ (a_n), $ so $ a_1 = 2001, $ $ a_2 = 2002, $ and $ a_3 = 2003, $ and \[a_n = a_{n - 2} + a_{n - 3} - a_{n - 1}.\]We can write this as \[a_n - a_{n - 2} = a_{n - 3} - a_{n - 1} = -(a_{n - 1} - a_{n - 3}).\]Let $ b_n = a_n - a_{n - 2}, $ so \[b_n = -b_{n - 1}.\]Also, $ b_{n - 1} = -b_{n - 2}, $ so $ b_n = b_{n - 2} $. Since $ b_4 = 2000 - 2002 = -2, $ it follows that $ b_n = -2 $ for all even $ n \ge 4 $. Then $ a_n - a_{n - 2} = -2 $ for all even $ n \ge 4 $. This means \[a_2, a_4, a_6, a_8, \dots\]is an arithmetic sequence with common difference $ -2 $. Hence, $ a_{2004} = 2002 - 1001 \cdot 2 = \boxed{0} $.