11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n, n \in \mathbf{N}^{*}$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}, T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, find the smallest positive integer $n$ such that $T_{n}>\frac{9}{20}$.
(1) From $2 S_{n}-n a_{n}=n$, we get $2 S_{n+1}-(n+1) a_{n+1}=n+1$. Subtracting the above two equations, we get
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1
$$

Thus,
$$
n a_{n}-(n-1) a_{n+1}=1
$$

Therefore,
$$
(n+1) a_{n+1}-n a_{n+2}=1
$$
(1) - (2), we get
$$
n a_{n}-2 n a_{n+1}+n a_{n+2}=0
$$

So $a_{n}+a_{n+2}=2 a_{n+1}$, which means the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence.
From $2 S_{1}-a_{1}=1$ and $a_{2}=3$, we get $a_{1}=1$, and the common difference $d=2$, so $a_{n}=1+2(n-1)=2 n-1$; (10 points)
(2) From (1), we know $b_{n}=\frac{1}{(2 n-1) \sqrt{2 n+1}+(2 n+1) \sqrt{2 n-1}}$, so
$$
b_{n}=\frac{1}{\sqrt{2 n-1} \cdot \sqrt{2 n+1} \cdot(\sqrt{2 n-1}+\sqrt{2 n+1})}=\frac{1}{2} \times \frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{\sqrt{2 n-1} \cdot \sqrt{2 n+1}}=\frac{1}{2}\left(\frac{1}{\sqrt{2 n-1}}-\frac{1}{\sqrt{2 n+1}}\right)
$$

Thus,
$$
T_{n}=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\cdots+\frac{1}{2}\left(\frac{1}{\sqrt{2 n-1}}-\frac{1}{\sqrt{2 n+1}}\right)=\frac{1}{2}\left(1-\frac{1}{\sqrt{2 n+1}}\right)
$$

From $T_{n}>\frac{9}{20}$, we get $\frac{1}{2}\left(1-\frac{1}{\sqrt{2 n+1}}\right)>\frac{9}{20}$, so
$$
\frac{1}{\sqrt{2 n+1}}<\frac{1}{10}, n > \frac{99}{2}
$$

Therefore, the smallest positive integer $n$ for which $T_{n}>\frac{9}{20}$ holds is 50.
(20 points)