We state that $a_{n}=\frac{(n-1)!!}{(n-2)!!}$, where $m!!=m\cdot (m-2)\cdot (m-4)\dots$ (all positive integer with the same parity as $m$ and not greater than $m$). Indeed, by induction: 1) for $n=3$ and $n=4$ statement is correct. 2) if $n>2$ then we have $a_{n+2}=\frac{1}{a_{n+1}}+a_n=\frac{(n-1)!!}{n!!}+\frac{(n-1)!!}{(n-2)!!}=\frac{(n+1)!!}{n!!}$. So $a_{2004}=\frac{2003!!}{2002!!}=\frac{2003!!\cdot 2002!!}{2002!!\cdot 2002!!}=\frac{2003!}{2^{2002}\cdot 1001!^2}$.

It is very interesting! Since we have $a_{n+2}a_{n+1}-a_{n+1}a_{n}=1, $ the sequence $a_{n+1}a_n $ is arithmetic progression with comon difference 1. so we obtain $a_{n+1}a_{n}=a_{2}a_{1}+1\cdot(n-1).$ i.e. $a_{n}a_{n+1}=n$ Substituting this recursive formula for $n=1,3,5,7,\cdot\cdot\cdot,2001,2003$ we obtain $a_{1}a_{2}=1,a_{3}a_{4}=3,a_{5}a_{6}=5,a_{7}a_{8}=7,\cdot\cdot\cdot,a_{2001}a_{2002}=2001,a_{2003}a_{2004}=2003$,respectively. Multiplying both sides of each equatios,we have $a_{1}a_{2}a_{3}\cdot\cdot\cdot a_{2003}a_{2004}=1\cdot3\cdot5\cdot\cdot\cdot\cdot 2001\cdot2003$ 　　　　　　　　　　　　　　　　① Similaly,substituting for $n=2,4,6,\cdot\cdot\cdot,2000,2002$ we obtain $a_{2}a_{3}=2,a_{4}a_{5}=4,a_{6}a_{7}=6,\cdot\cdot\cdot, a_{2000}a_{2001}=2000,a_{2002}a_{2003}=2002,$ respctively. Therefore we obtain $a_{2}a_{3}a_{4}\cdot\cdot\cdot ,a_{2002}a_{2003}=2\cdot4\cdot6\cdot\cdot\cdot2000\cdot2002$　　　　　　　　　　　　　　　　　② Dividing ①　ｂｙ　②　of both sides,we yield ,from $a_{1}=1$, $a_{2004}=\frac{1\cdot3\cdot5\cdot\cdot\cdot 2001\cdot2003}{2\cdot4\cdot6\cdot\cdot\cdot 2000\cdot2002}$. Thank you kunny

Thank you for confirmation of my answer

We just have this recursion obtained $a_{n+1}a_{n}=a_{n}a_{n-1}$ which when combined with the initial values just gives $a_{n}=\frac{n-1}{a_{n-1}}$ which now just gives $a_{n}=\frac{(n-1)!!}{(n-2)!!}$ and so we are done,I don't simplify this further.