11. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=\frac{9}{4}, \\
2 a_{n+1} a_{n}-7 a_{n+1}-3 a_{n}+12=0(n \geqslant 1) .
\end{array}
$$
(1) Let $c_{n}=a_{n}-2$, find the general term formula for the sequence $\left\{c_{n}\right\}$;
(2) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $b_{n}=\frac{n^{2}}{n+1} a_{n}$. Find the maximum positive integer $n$ such that $\sum_{k=1}^{n}\left[b_{k}\right] \leqslant 2019$.
(1) From the given condition, we have $a_{n}=c_{n}+2$.
Substituting into the recursive formula, we get
$$
\begin{array}{l}
2\left(c_{n+1}+2\right)\left(c_{n}+2\right)-7\left(c_{n+1}+2\right)-3\left(c_{n}+2\right)+12 \\
=0 \\
\Rightarrow 2 c_{n+1} c_{n}-3 c_{n+1}+c_{n}=0 \\
\Rightarrow \frac{1}{c_{n+1}}-1=3\left(\frac{1}{c_{n}}-1\right) .
\end{array}
$$

Thus, $\left\{\frac{1}{c_{n}}-1\right\}$ is a geometric sequence with the first term $\frac{1}{c_{1}}-1=3$ and common ratio 3.
Therefore, $\frac{1}{c_{n}}-1=3^{n} \Rightarrow c_{n}=\frac{1}{3^{n}+1}$.
(2) From (1), we know
$$
\begin{array}{l} 
a_{n}=c_{n}+2=\frac{1}{3^{n}+1}+2, \\
b_{n}=\frac{n^{2}}{n+1} a_{n}=\frac{n^{2}}{n+1}\left(\frac{1}{3^{n}+1}+2\right) \\
= \frac{n^{2}}{(n+1)\left(3^{n}+1\right)}+\frac{2 n^{2}}{n+1} \\
= 2(n-1)+\frac{2}{n+1}+\frac{n^{2}}{(n+1)\left(3^{n}+1\right)} \\
=2(n-1)+\frac{2 \times 3^{n}+n^{2}+2}{(n+1)\left(3^{n}+1\right)} .
\end{array}
$$

When $n \geqslant 2$,
$$
\begin{array}{l}
3^{n}=(1+2)^{n}=1+\mathrm{C}_{n}^{1} \times 2+\mathrm{C}_{n}^{2} \times 2^{2}+\cdots \\
>1+2 n, \\
(n+1)\left(3^{n}+1\right)-\left(2 \times 3^{n}+n^{2}+2\right) \\
=(n-1) \times 3^{n}-n^{2}+n-1 \\
>(n-1)(1+2 n)-n^{2}+n-1 \\
=n^{2}-2>0 .
\end{array}
$$

Therefore, when $n \geqslant 2$,
$$
\begin{array}{l}
0<\frac{2 \times 3^{n}+n^{2}+2}{(n+1)\left(3^{n}+1\right)}<1 \\
\Rightarrow\left[b_{n}\right]=2(n-1) .
\end{array}
$$

When $n=1$, $b_{1}=\frac{6+1+2}{2 \times 4}=\frac{9}{8}$, then
$$
\left[b_{1}\right]=1 \text {. }
$$

Thus, when $n \geqslant 2$,
$$
\begin{array}{l}
\sum_{k=1}^{n}\left[b_{k}\right]=1+\sum_{k=2}^{n} 2(k-1) \\
=1+n(n-1)=n^{2}-n+1 .
\end{array}
$$

From $n^{2}-n+1 \leqslant 2019\left(n \in \mathbf{Z}_{+}\right)$
$$
\Rightarrow n \leqslant 45 \text {. }
$$

Therefore, the required value is 45.