11. Find all positive integer tuples $(x, y, u, v)$ that satisfy $\sqrt{x y}+\sqrt{u v}$ is a rational number, and $\left|\frac{x}{9}-\frac{y}{4}\right|=\left|\frac{u}{3}-\frac{v}{12}\right|=u v-x y$.
If $x y$ is not a perfect square, let $\sqrt{x y}=a \sqrt{b}\left(a, b \in \mathbf{N}^{*}\right.$, and $\sqrt{b}$ is irreducible).
(1) If $u v$ is a perfect square, then $\sqrt{x y} \notin \mathbf{Q}, \sqrt{u v} \in \mathbf{Q} \Rightarrow \sqrt{x y}+\sqrt{u v} \notin \mathbf{Q}$, which does not meet the requirement;
(2) If $u v$ is not a perfect square, let $\sqrt{u v}=c \sqrt{d}\left(c, d \in \mathbf{N}^{*}\right.$, and $\sqrt{d}$ is irreducible).

Then $\sqrt{x y}+\sqrt{u v}=a \sqrt{b}+c \sqrt{d} \notin \mathbf{Q}$, which does not meet the requirement.
Therefore, $x y$ is a perfect square, and similarly, $u v$ is also a perfect square.
From $\left|\frac{x}{9}-\frac{y}{4}\right|=\left|\frac{u}{3}-\frac{v}{12}\right|=u v-x y \Rightarrow \frac{x}{9}-\frac{y}{4} \in \mathbf{N}, \frac{u}{3}-\frac{v}{12} \in \mathbf{N} \Rightarrow 36|4 x-9 y, 36| 12 u-3 v$,
Then $4 x-9 y \equiv 4 x \equiv 0(\bmod 9) \Rightarrow 9|x ; 4 x-9 y \equiv-9 y \equiv 0(\bmod 4) \Rightarrow 4| y$;
$12 u-3 v \equiv-3 v \equiv 0(\bmod 4) \Rightarrow 4 \mid v$;
(1) When $3 \mid u$, $\left.\frac{u}{3}-\frac{v}{12} \in \mathbf{N} \Rightarrow 12 \right\rvert\, v$, let $\frac{x}{9}=a, \frac{y}{4}=b, \frac{u}{3}=c, \frac{v}{12}=d$, then $a, b, c, d \in \mathbf{N}^{*}$,

and $|a-b|=|c-d|=36(c d-a b)$. Without loss of generality, assume $a \geqslant b, c \geqslant d$,
Let $s=c d-a b \Rightarrow a-b=c-d=36(c d-a b)=36 s, s \in \mathbf{N}$,
Then $36 s=36[d(36 s+d)-b(36 s+b)]$
$\Rightarrow s=36 s d-36 s b+d^{2}-b^{2}=(d-b)(36 s+d+b)$.
If $d \neq d \Rightarrow d-b \geqslant 1 \Rightarrow s \geqslant 36 s+d+b \Rightarrow 35 s+d+b \leqslant 0$, contradiction.
Thus $d=b \Rightarrow s=0 \Rightarrow a=b, c=d \Rightarrow a=b=c=d$,
So $x=9 k, y=4 k, u=3 k, v=12 k, k \in \mathbf{N}^{*}$.
(2) When $3 \nmid u$, $\left.\frac{u}{3}-\frac{v}{12} \in \mathbf{N} \Rightarrow 12 \right\rvert\, v$, let $\frac{x}{9}=a, \frac{y}{4}=b, \frac{u}{3}=c$,

Let $v=4 w \Rightarrow \frac{v}{12}=\frac{w}{3}=d$, then $a, b, 3 c, 3 d \in \mathbf{N}^{*}$, and $|a-b|=|c-d|=36(c d-a b)$.
Without loss of generality, assume $a \geqslant b, u \geqslant w$, i.e., $a-b=\frac{u}{3}-\frac{w}{3}=4 u w-36 a b$.
Let $t=u w-9 a b \Rightarrow a-b=4 t, u-w=12 t, t \in \mathbf{N}^{*}$.
Then $4 t=4 w(12 t+w)-36 b(4 t+b) \Rightarrow t=w(12 t+w)-9 b(4 t+b)$
$=12 t w-36 t b+w^{2}-9 b^{2}=(w-3 b)(12 t+w+3 b)$.
If $w \neq 3 b \Rightarrow w-3 b \geqslant 1 \Rightarrow t \geqslant 12 t+w+3 b \Rightarrow 11 t+w+3 b \leqslant 0$, contradiction.
Thus $w=3 b \Rightarrow t=0 \Rightarrow a=b, c=d \Rightarrow a=b=c=d$,
So $x=9 k, y=4 k, u=3 k, v=12 k, k \in \mathbf{N}^{*}$.
In summary, all the positive integer solutions $(x, y, u, v)=(9 k, 4 k, 3 k, 12 k), k \in \mathbf{N}^{*}$.