11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{9}{4}, 2 a_{n+1} a_{n}-7 a_{n+1}-3 a_{n}+12=0\left(n \in \mathbf{N}_{+}\right)$.
(1) Let $c_{n}=a_{n}-2$, find the general term formula of the sequence $\left\{c_{n}\right\}$;
(2) Let $b_{n}=\frac{n^{2}}{n+1} a_{n}$, find the maximum positive integer $n$ such that $\left[b_{1}\right]+\left[b_{2}\right]+\cdots+\left[b_{n}\right] \leqslant 2019$.
(Here, the symbol $[x]$ represents the greatest integer not exceeding $x$)
$$
\begin{array}{l}
\text { (1) } 2\left(c_{n+1}+2\right)\left(c_{n}+2\right)-7\left(c_{n+1}+2\right)-3\left(c_{n}+2\right)+12 \\
=2\left(c_{n+1} c_{n}+2 c_{n+1}+2 c_{n}\right)-7 c_{n+1}-3 c_{n}=2 c_{n+1} c_{n}-3 c_{n+1}+c_{n}=0 \\
\Rightarrow \frac{1}{c_{n+1}}=\frac{3}{c_{n}}-2 \Rightarrow \frac{1}{c_{n+1}}-1=3\left(\frac{1}{c_{n}}-1\right) \text {, and } \frac{1}{c_{1}}-1=3,
\end{array}
$$

Therefore, $\frac{1}{c_{n}}-1=3^{n} \Rightarrow c_{n}=\frac{1}{3^{n}+1}\left(n \in \mathbf{N}_{+}\right)$.
(2) From (1), we know $a_{n}=\frac{1}{3^{n}+1}+2 \Rightarrow b_{n}=\frac{n^{2}}{(n+1)\left(3^{n}+1\right)}+\frac{2 n^{2}}{n+1}$
$$
=\frac{n^{2}}{(n+1)\left(3^{n}+1\right)}+\frac{2}{n+1}+2(n-1) \text {. }
$$

Thus, $\left[b_{1}\right]=1,\left[b_{2}\right]=2,\left[b_{3}\right]=4,\left[b_{4}\right]=6$, for $n \geqslant 2$, $\left[b_{n}\right]=2(n-1)$.
Therefore, $\left[b_{1}\right]+\left[b_{2}\right]+\cdots+\left[b_{n}\right]=n^{2}-n+1 \leqslant 2019 \Rightarrow n^{2}-n-2018 \leqslant 0$, since $45^{2}=2025,46^{2}=2116$, the largest positive integer $n$ that satisfies the condition is $n=45$.