6. If the sum of the areas of three square pieces of paper with integer side lengths is 2004, and the area of the largest square piece of paper is $S_{1}$, and the area of the smallest square piece of paper is $S_{2}$, then the maximum value of $\frac{S_{1}}{S_{2}}$ is $\qquad$
6.484.

Let the side lengths of the squares be $a$, $b$, and $c$, with $a<b<c$. Then, $a^{2}+b^{2}+c^{2}=2004$.
Obviously, $S_{2}=a^{2}, S_{1}=c^{2} \Rightarrow \frac{S_{1}}{S_{2}}=\frac{c^{2}}{a^{2}}$.
It is easy to see that $40^{2}+20^{2}+2^{2}=2004$, $44^{2}+8^{2}+2^{2}=2004$.

To maximize $\frac{S_{1}}{S_{2}}$, $S_{1}$ should be as large as possible, and $S_{2}$ should be as small as possible.
Thus, $c^{2}$ is maximized at $44^{2}=1936$, and $a^{2}$ is minimized at $2^{2}=4$, at which point, $b^{2}=8^{2}=64$.
Therefore, the maximum value of $\frac{S_{1}}{S_{2}}$ is $\frac{1936}{4}=484$.