14. The sequence $\left\{a_{n}\right\}$ is an arithmetic sequence, and it satisfies $3 a_{5}=8 a_{12}>0$. The sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=a_{n} \cdot a_{n+1} \cdot a_{n+2}\left(n \in \mathbf{N}^{*}\right)$, and the sum of the first $n$ terms of $\left\{b_{n}\right\}$ is denoted as $S_{n}$. For what value of $n$ does $S_{n}$ achieve its maximum value? Explain your reasoning.
Given that $3 a_{5}=8 a_{12}>0$, we have $3 a_{5}=8\left(a_{5}+7 d\right)$, solving for $a_{5}$ gives $a_{5}=-\frac{56}{5} d>0$, so $d<0$, and $a_{1}>a_{2}>a_{3}>\cdots>a_{16}>0>a_{17}>a_{18}>\cdots$

Therefore,
$$
b_{1}>b_{2}>b_{3}>\cdots>b_{14}>0>b_{17}>b_{18}>\cdots
$$

And
$$
b_{15}=a_{15} a_{16} a_{17}<0, \quad b_{16}=a_{16} a_{17} a_{18}>0
$$

So
$$
S_{14}>S_{13}>\cdots>S_{1}, \quad S_{14}>S_{15}, \quad S_{15}<S_{16}, \quad S_{16}>S_{17}>S_{18}>\cdots
$$

Also,
$$
S_{16}-S_{14}=b_{15}+b_{16}=a_{16} a_{17}\left(a_{15}+a_{18}\right)=a_{16} a_{17}\left(-\frac{6}{5} d+\frac{9}{5} d\right)=\frac{3}{5} d a_{16} a_{17}>0
$$

Therefore, $S_{n}$ is maximized at $S_{16}$, i.e., $n=16$ when $S_{n}$ reaches its maximum value.