II. (50 points) Find all positive integers $x, y, z$ such that $\sqrt{\frac{2009}{x+y}}+\sqrt{\frac{2009}{y+z}}+\sqrt{\frac{2009}{z+x}}$ is an integer.
II. First, prove a lemma.
Lemma If $p, q, r$ are rational numbers, and $S=\sqrt{p}+\sqrt{q}+\sqrt{r}$ is also a rational number, then $\sqrt{p}, \sqrt{q}, \sqrt{r}$ must all be rational numbers.
Proof of the lemma: Notice that
$(\sqrt{p}+\sqrt{q})^{2}=(S-\sqrt{r})^{2}$
$\Rightarrow 2 \sqrt{p q}=S^{2}+r-p-q-2 S \sqrt{r}$.
Let $M=S^{2}+r-p-q>0$.
Then $2 \sqrt{p q}=M-2 S \sqrt{r}$.
Squaring both sides of the above equation gives
$4 p q=M^{2}+4 S^{2} r-4 M S \sqrt{r}$.
Thus, $\sqrt{r}=\frac{M^{2}+4 S^{2} r-4 p q}{4 M S}$ is a rational number.
Similarly, $\sqrt{p}, \sqrt{q}$ are also rational numbers.
Now, prove the original problem.
Assume $x, y, z$ are positive integers that satisfy the conditions.
Also, $\sqrt{\frac{2009}{x+y}}+\sqrt{\frac{2009}{y+z}}+\sqrt{\frac{2009}{x+z}}$ is an integer. By the lemma, $\sqrt{\frac{2009}{x+y}}, \sqrt{\frac{2009}{y+z}}, \sqrt{\frac{2009}{x+z}}$ are rational numbers.
Let $\sqrt{\frac{2009}{x+y}}=\frac{a}{b}$, and $(a, b)=1$,
$\sqrt{\frac{2009}{y+z}}=\frac{a}{c}, \sqrt{\frac{2009}{x+z}}=\frac{a}{d}$.
Then $2009 b^{2}=(x+y) a^{2}$.
Therefore, $a^{2}$ divides 2009.
So, $a=1$.
Thus, $x+y=2009 b^{2}$.
Similarly, $y+z=2009 c^{2}$,
$z+x=2009 d^{2}$.
(2)
Substituting equations (1), (2), and (3) into the original expression, we know that $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is a positive integer.
(1) + (3) gives
$2 x+y+z=2009 b^{2}+2009 d^{2}$.
Substituting equation (2) into the above equation gives
$x=\frac{2009}{2}\left(b^{2}+d^{2}-c^{2}\right)$.
Similarly, $y=\frac{2009}{2}\left(b^{2}+c^{2}-d^{2}\right)$,
$z=\frac{2009}{2}\left(c^{2}+d^{2}-b^{2}\right)$.
Notice that $\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leqslant 3$, and consider the following three cases.
(1) When $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=3$, $b=c=d=1$.
From equations (1), (2), and (3), we have $x+y=y+z=z+x=2009$ with no positive integer solutions.
(2) When $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=2$, one of $b, c, d$ must be 1, and the other two must be 2. In this case, there are no $x, y, z$ that satisfy the conditions.
(3) When $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$, assume without loss of generality that
$b \geqslant c \geqslant d>1$.
Then $\frac{3}{d} \geqslant \frac{1}{b}+\frac{1}{c}+\frac{1}{d}$. Hence, $d=2$ or $d=3$.
(i) If $d=3$, then $b=c=3$. In this case, there are no $x, y, z$ that satisfy the conditions.
(ii) If $d=2$, then $c>2$, and
$\frac{2}{c} \geqslant \frac{1}{b}+\frac{1}{c}=\frac{1}{2}$.
Thus, $c=3$ or $c=4$.
a) If $c=3$, then $b=6$. In this case, there are no $x, y, z$ that satisfy the conditions.
b) If $c=4$, then $b=4$. In this case, there exist $x, y, z$ that satisfy the conditions. Calculating, we get
$$
\begin{array}{l}
x=\frac{2009}{2}(16+4-16)=4018, \\
y=\frac{2009}{2}(16+16-4)=28126, \\
z=\frac{2009}{2}(16+4-16)=4018 .
\end{array}
$$

Therefore, one of $x, y, z$ is 28126, and the other two
are both 4018.