The answer is $1003$.

Note that
$$r = \frac{2004 + \sqrt{2004^2 + 4}}{2}$$
is an irrational number greater than $1$. Since $a_n \in \mathbb{Z}^+$, we have $ra_n \notin \mathbb{Q}$. Thus, we obtain
$$a_{n+1} < ra_n < a_{n+1} + 1$$
for any $n \ge 0$. Equivalently, we have
$$a_n - \frac{1}{r} < \frac{a_{n+1}}{r} < a_n.$$
As $r > 1$, this yields $a_n - 1 < \frac{a_{n+1}}{r} < a_n$, and hence
$$\left[ \frac{a_{n+1}}{r} \right] = a_n - 1.$$
Next, observe that $r = 2004 + \frac{1}{r}$. It follows that
$$a_{n+1} = [ra_n] = \left[ 2004a_n + \frac{a_n}{r} \right] = 2004a_n + a_{n-1} - 1$$
since $2004a_n \in \mathbb{Z}$ and $\left[ \frac{a_n}{r} \right] = a_{n-1} - 1$ from above. Thus, we obtain
$$a_{n+1} \equiv a_{n-1} - 1 \pmod{2004}$$
for any $n \in \mathbb{Z}^+$. Then we can easily obtain
$$a_{2004} \equiv a_{2002} - 1 \equiv a_{2000} - 2 \equiv \cdots \equiv a_0 - 1002 \equiv 1003 \pmod{2004}.$$