China Western Mathematical Olympiad

According to the assumption we have
$$a_{n+2} a_{n+1} - a_{n+1} a_n = 1.$$
Thus, $\{a_{n+1} a_n\}$ is an arithmetic progression with first term $1$ and common difference $1$. Hence
$$a_{n+1} a_n = n, \quad n = 1, 2, \dots$$
So $a_{n+2} = \frac{n+1}{a_{n+1}} = \frac{n+1}{n} = \frac{n+1}{n} a_n, \quad n = 1, 2, \dots$. Consequently,
$$\begin{align*} a_{2004} &= \frac{2003}{2002} a_{2002} = \frac{2003}{2002} \cdot \frac{2001}{2000} a_{2000} \\ &= \dots = \frac{2003}{2002} \cdot \frac{2001}{2000} \cdot \dots \cdot \frac{3}{2} a_2 \\ &= \frac{3 \cdot 5 \cdot \dots \cdot 2003}{2 \cdot 4 \cdot \dots \cdot 2002} \end{align*}$$