Solution:
There exists. Let us take $a_{0}=1$, $a_{i}=2004!a_{0} a_{1} \ldots a_{i-1}-1$, $i \geq 1$ and $A_{i}=\{2,3, \ldots, 2004, a_{0}, a_{1}, \ldots, a_{i}\}$, $i \geq 0$. Then
$$\begin{gathered} \left(\prod_{a \in A_{i-1}} a-\sum_{a \in A_{i-1}} a^{2}\right)-\left(\prod_{a \in A_{i}} a-\sum_{a \in A_{i}} a^{2}\right)-1 \\ \quad=a_{i}^{2}-1-\left(a_{i}-1\right) \prod_{a \in A_{i-1}} a \\ \quad=\left(a_{i}-1\right)\left(a_{i}+1-2004!a_{0} a_{1} \ldots a_{i-1}\right)=0 \end{gathered}$$
Hence
$$\prod_{a \in A_{n}} a=\sum_{a \in A_{n}} a^{2} \text{ for } n=\prod_{a \in A_{0}} a-\sum_{a \in A_{0}} a^{2}$$