$\qquad$Suppose $n$ is a given positive integer, $D_{n}$ is the set of all positive divisors of $2^{n} 3^{n} 5^{n}$, $S \subseteq D_{n}$, and no number in $S$ can divide another number in $S$. Find the maximum value of $|S|$.
Parse $|S|_{\max }=\left[\frac{3(n+1)^{2}+1}{4}\right]$.
Obviously, every number in $D_{n}$ has the form $2^{\alpha} 3^{\beta} 5^{\gamma}$, where $0 \leqslant \alpha, \beta, \gamma \leqslant n$. Below, we use the array $(\alpha, \beta, \gamma)$ to represent the number $2^{\alpha} 3^{\beta} 5^{\gamma}$.
Consider the following sets:
$$
\begin{array}{l}
A_{i, j}-\{(i, j, \alpha), 0 \leqslant \alpha \leqslant n-j\} \cup \\
\{(i, \beta, n-j), j \leqslant \beta \leqslant n\}, \\
i=0,1,2, \cdots, n, j=0,1,2, \cdots,\left[\frac{n}{2}\right] ; \\
B_{i, j}=\{(k, i, j), k=0,1,2, \cdots, n\}, \\
i=\left[\frac{n}{2}\right]+1, \cdots, n, j=0,1,2, \cdots, n-\left[\frac{n}{2}\right]-1 . \\
\end{array}
$$

The above has a total of

or
$$
\begin{aligned}
& \left(\left[\frac{n}{2}\right]+1\right) \cdot(n+1)+\left(n-\left[\frac{n}{2}\right]\right)^{2} \\
= & \left(\frac{n}{2}+1\right)(n+1)+\frac{n^{2}}{4}=\frac{3(n+1)^{2}+1}{4} \text { ( } n \text { is even), }
\end{aligned}
$$

or $\begin{array}{l}\left(\left[\frac{n}{2}\right]+1\right)(n+1) \\ =\left(\frac{n-1}{2}+1\right)(n+1) \\ = \frac{3(n+1)^{2}}{4}(n \text { is odd } \\ \text { In summary, there are }\left[\frac{3(n+1)^{2}+1}{4}\right] \text { sets. }\end{array}$
Since at most one number in each set belongs to $S$ (because any two numbers in each of the above sets have a multiple relationship), we get that the above sets are mutually exclusive and include all the numbers, i.e., $\cup A_{i, j} \cup B_{i, j}=D_{n}$. Therefore, we can get $|S| \leqslant\left[\frac{3(n+1)^{2}+1}{4}\right]$.

On the other hand, consider the number of arrays $(\alpha, \beta, \gamma)$ that satisfy $\alpha+\beta+\gamma=n+t\left(\right.$ where $\left.t=\left[\frac{n}{2}\right]\right)$.
By the condition $0 \leqslant \alpha, \beta, \gamma \leqslant n$, we get
$$
\begin{aligned}
& (n+t+1)+(n+t)+\cdots+1-3[t+(t-1)+\cdots+1] \\
= & \frac{1}{2}(n+t+2)(n+t+1)-\frac{3}{2} t(t+1) \\
= & \frac{1}{2}(n+1)(n+2)+\frac{1}{4} n^{2}-\left(t-\frac{n}{2}\right)^{2} \\
= & \frac{1}{4}\left[3(n+1)^{2}+1\right]-\left(t-\frac{n}{2}\right)^{2} \\
= & {\left[\frac{3(n+1)^{2}+1}{4}\right] \cdot\left(\text { where } t=\left[\frac{n}{2}\right]\right) }
\end{aligned}
$$
(When one of $\alpha, \beta, \gamma$ is greater than $n$, there are $3(t+(t-1)+\cdots+1)$ sets of solutions, so they should be removed)
Take $S=\left\{2^{\alpha} 3^{\beta} 5^{\gamma} \left\lvert\, \alpha+\beta+\gamma=n+\left[\frac{n}{2}\right]\right., 0 \leqslant \alpha, \beta, \gamma \leqslant n\right\}$, from the above we get $|S|=$ $\left[\frac{3(n+1)^{2}+1}{4}\right]$, and any two numbers in $S$ do not have a multiple relationship, otherwise, assume $2^{\alpha_{1}} 3^{\beta_{1}} 5^{\gamma_{1}} \mid 2^{\alpha_{2}} 3^{\beta_{2}} 5^{\gamma_{2}}$, then $\alpha_{1} \leqslant \alpha_{2}, \beta_{1} \leqslant \beta_{2}, \gamma_{1} \leqslant \gamma_{2}, \alpha_{1}+\beta_{1}+\gamma_{1}=\alpha_{2}+\beta_{2}+\gamma_{2}$, therefore $\alpha_{1}=\alpha_{2}, \beta_{1}=\beta_{2}, \gamma_{1}=\gamma_{2}$, which leads to a contradiction!
In summary, $|S|_{\max }=\left[\frac{3(n+1)^{2}+1}{4}\right]$.