6. Given $\sqrt{2009}=\sqrt{x}+\sqrt{y}$, and $0<x<y$. Find all integer pairs $(x, y)$ that satisfy the equation.
From $\sqrt{2009}=\sqrt{x}+\sqrt{y}$, we get
$$
\sqrt{\frac{x}{2009}}+\sqrt{\frac{y}{2009}}=1 \text {. }
$$

It is easy to prove that $\sqrt{\frac{x}{2009}}$ and $\sqrt{\frac{y}{2009}}$ are both rational numbers.
Let $\sqrt{\frac{x}{2009}}=\frac{q_{1}}{p_{1}}\left(p_{1} 、 q_{1}\right.$ are both positive integers, and $p_{1}$ and $q_{1}$ are coprime, $q_{1}<p_{1}$ ).
$$
\begin{array}{l}
\text { Then } p_{1}^{2} x=2009 q_{1}^{2} \Rightarrow p_{1}^{2} \mid 2009 q_{1}^{2} \\
\Rightarrow p_{1}^{2} \mid 2009 \text {. }
\end{array}
$$

Since $2009=7^{2} \times 41$, and $p_{1} \geqslant 2$, thus $p_{1}=7$.
Therefore, $x=41 q_{1}^{2}\left(1 \leqslant q_{1} \leqslant 6\right)$.
Similarly, $y=41 q_{2}^{2}\left(1 \leqslant q_{2} \leqslant 6\right)$.
$$
\begin{array}{l}
\text { From } \sqrt{x}+\sqrt{y}=\sqrt{2009} \\
\Rightarrow q_{1} \sqrt{41}+q_{2} \sqrt{41}=7 \sqrt{41} \\
\Rightarrow q_{1}+q_{2}=7 .
\end{array}
$$

Also, $0<x<y$, so
$q_{1}<q_{2} \Rightarrow q_{1}$ can only be $1,2,3$
$$
\Rightarrow\left\{\begin{array}{l}
x=41 \times 1^{2}, \\
y=41 \times 6^{2}
\end{array} ;\left\{\begin{array}{l}
x=41 \times 2^{2}, \\
y=41 \times 5^{2}
\end{array} ;\left\{\begin{array}{l}
x=41 \times 3^{2}, \\
y=41 \times 4^{2}
\end{array}\right.\right.\right.
$$