Four. (30 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=1+\frac{1}{a_{n}}\left(n \in \mathbf{N}_{+}\right)$, and for any positive integer $n(n \geqslant 2)$, the inequality
$$
\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}
$$

always holds, find the maximum value of the integer $m$.
(1) From
$$
\begin{array}{l}
a_{n}=2 a_{n} a_{n+1}+3 a_{n+1} \\
\Rightarrow a_{n+1}=\frac{a_{n}}{2 a_{n}+3} \Rightarrow a_{n}>0 . \\
\text { Hence } \frac{1}{a_{n+1}}-\frac{3}{a_{n}}=2 \Rightarrow \frac{1}{a_{n+1}}+1=3\left(\frac{1}{a_{n}}+1\right) .
\end{array}
$$

Therefore, $\left\{\frac{1}{a_{n}}+1\right\}$ is a geometric sequence with the first term 3 and common ratio 3.
Thus, $\frac{1}{a_{n}}+1=3^{n} \Rightarrow a_{n}=\frac{1}{3^{n}-1}\left(n \in \mathbf{N}_{+}\right)$.
(2) From (1), we get $b_{n}=1+\frac{1}{a_{n}}=3^{n}$.
Therefore, $\log _{3} b_{k}=k(k=1,2, \cdots)$.
Hence, $\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}$
$\Leftrightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}>\frac{m}{24}$.
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}$. Then
$$
\begin{array}{l}
f(n+1)-f(n) \\
=\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1} \\
=\frac{1}{2 n+1}-\frac{1}{2 n+2}>0 .
\end{array}
$$

Therefore, $f(n)$ is monotonically increasing.
Hence, $f(n)_{\min }=f(2)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$.
Thus, $\frac{7}{12}>\frac{m}{24} \Rightarrow m<14$.
Therefore, the maximum integer value of $m$ is 13.