(1) From $2 S_{n}-n a_{n}=n$, we get
$$
2 S_{n+1}-(n+1) a_{n+1}=n+1 \text {. }
$$

Subtracting the two equations yields
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \text {. }
$$

Thus, $n a_{n}-(n-1) a_{n+1}=1$,
$$
(n+1) a_{n+1}-n a_{n+2}=1 \text {. }
$$
Subtracting (2) from (1) and rearranging gives
$$
a_{n}+a_{n+2}=2 a_{n+1} \text {. }
$$

Hence, $\left\{a_{n}\right\}$ is an arithmetic sequence.
From $2 S_{1}-a_{1}=1$ and $a_{2}=3$, we get $a_{1}=1$.
Thus, the common difference of the sequence $\left\{a_{n}\right\}$ is $d=2$.
Therefore, $a_{n}=1+2(n-1)=2 n-1$.
(2) From (1), we know
$$
\begin{aligned}
b_{n} & =\frac{1}{(2 n-1) \sqrt{2 n+1}+(2 n+1) \sqrt{2 n-1}} \\
& =\frac{1}{2} \times \frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{\sqrt{2 n-1} \sqrt{2 n+1}} \\
& =\frac{1}{2}\left(\frac{1}{\sqrt{2 n-1}}-\frac{1}{\sqrt{2 n+1}}\right) .
\end{aligned}
$$

Thus, $T_{n}=\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{\sqrt{2 k-1}}-\frac{1}{\sqrt{2 k+1}}\right)$
$$
\begin{array}{l}
=\frac{1}{2}\left(1-\frac{1}{\sqrt{2 n+1}}\right)>\frac{9}{20} \\
\Rightarrow n \geqslant \frac{99}{2} .
\end{array}
$$

Therefore, the smallest positive integer $n$ that satisfies the condition is $n=50$.
11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n\left(n \in \mathbf{Z}_{+}\right)$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$; for the sequence $\left\{b_{n}\right\}$, find the sum of the first $n$ terms, and determine the smallest positive integer $n$ such that $T_{n}>\frac{9}{20}$.