Let $S \subseteq D_{n}$, and any number in $S$ cannot divide another number in $S$. Find the maximum value of $|S|$.
Consider the "two-dimensional" case first: $C_{n}=\left\{2^{\alpha} \cdot 3^{\beta} \mid 0 \leqslant \alpha \leqslant n, 0 \leqslant \beta \leqslant n\right\}$, $T \subseteq C_{n}$, and no number in $T$ can divide another number in $T$. Find the maximum value of $|T|$.

The numbers in $C_{n}$ can be represented by the points in the square $I=\{(\alpha, \beta) \mid 0 \leqslant \alpha, \beta \leqslant n$, and $\alpha, \beta$ are integers $\}$. $T$ can be a subset of $I$. On each line parallel to the $y$-axis, $T$ can have at most one point; on each line parallel to the $x$-axis, $T$ can also have at most one point. Therefore, $|T| \leqslant n + 1$.

On the other hand, the set $T$ consisting of the $n+1$ integer points on the line $x+y=n$ (1) satisfies the requirements. Therefore, the maximum value of $|T|$ is $n+1$.

The numbers in $D_{n}$ can be represented by the points in the cube $M=\{(\alpha, \beta, \gamma) \mid 0 \leqslant \alpha, \beta, \gamma \leqslant n$, and $\alpha, \beta, \gamma$ are integers $\}$. $S$ is a subset of $M$, and on each line parallel to the coordinate axes, $S$ can have at most one point. Therefore, $|S| \leqslant (n+1)^{2}$. However, $(n+1)^{2}$ is an upper bound for $|S|$, not the maximum value. Similar to (1), we should consider the plane $x+y+z=\frac{3 n}{2}$ passing through the center $\left(\frac{n}{2}, \frac{n}{2}, \frac{n}{2}\right)$ of $M$. But $\frac{3 n}{2}$ is not an integer when $n$ is odd. Therefore, we should consider the plane $x+y+z=\left[\frac{3 n}{2}\right]=n+\left[\frac{n}{2}\right]$ with integer points. The set $B$ of these points satisfies the requirements: for any two points $(\alpha, \beta, \gamma)$ and $\left(\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime}\right)$ in $B$, if $\alpha>\alpha^{\prime}$, then $\beta<\beta^{\prime}$ or $\gamma<\gamma^{\prime}$ must hold, meaning that no number $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}$ in $B$ can divide another number.

At the same time, the cube can be divided into $(n+1) \times\left(\left[\frac{n}{2}\right]+1\right)$ L-shaped regions and $\left(n-\left[\frac{n}{2}\right]\right)^{2}$ lines. These L-shaped regions are
$$
\begin{array}{c}
\left\{(p, q, z) \left\lvert\, 0 \leqslant z \leqslant n+\left[\frac{n}{2}\right]-p-q\right.\right\} \cup\left\{\left.\left(p, y, n+\left[\frac{n}{2}\right]-p-q\right) \right\rvert\, q \leqslant y \leqslant n\right\}, \\
p=0,1, \cdots, n ; q=0,1, \cdots,\left[\frac{n}{2}\right] .
\end{array}
$$

These lines are
$$
\begin{array}{c}
\{(x, q, r) \mid 0 \leqslant x \leqslant n\}, \\
q-\left[\begin{array}{l}
n \\
2
\end{array}\right]+1, \cdots, n-1, n ; r=0,1, \cdots, n-\left[\frac{n}{2}\right]-1 .
\end{array}
$$

On each L-shaped region, $S$ can have at most one point. On each line, $S$ can also have at most one point. On each L-shaped region, there is a point $\left(p, q, n+\left[\frac{n}{2}\right]-p-q\right) \in M$. On each line, there is also a point $\left(n+\left[\frac{n}{2}\right]-q-r, q, r\right) \in M$ (since $\left[\frac{n}{2}\right]+1 \leqslant q+r \leqslant 2 n-\left[\frac{n}{2}\right]-1 \leqslant n+$ $\left[\frac{n}{2}\right]$, so
$$
\left.0 \leqslant n+\left[\frac{n}{2}\right]-q-r \leqslant n-1\right) .
$$

Thus, $|M|$ is the maximum value of $|S|$. This maximum value is
$$
\begin{aligned}
& (n+1)\left(\left[\frac{n}{2}\right]+1\right)+\left(n-\left[\frac{n}{2}\right]\right)^{2} \\
= & \left\{\begin{array}{l}
(n+1)\left(\frac{n}{2}+1\right)+\left(\frac{n}{2}\right)^{2}=\frac{3(n+1)^{2}+1}{4}, n \text { is even, } \\
(n+1) \cdot \frac{n+1}{2}+\left(\frac{n+1}{2}\right)^{2}=\frac{3(n+1)^{2}}{4}, n \text { is odd. }
\end{array}\right. \\
= & {\left[\frac{3(n+1)^{2}+1}{4}\right] . }
\end{aligned}
$$