Example 4 Given that $S$ is a subset of the set $\{1,2, \cdots, n\}$, $S$ contains no number that divides another, and no two numbers are coprime. Find the maximum value of $S$.
【Analysis】First, guess the maximum value $k$ by construction. To construct a set without divisibility relations, we can choose $\{m, m+1, \cdots, n\}\left(m>\frac{n}{2}\right)$. To construct a set of numbers that are pairwise not coprime, we can choose $\left\{2,4, \cdots, 2\left[\frac{n}{2}\right]\right\}([x]$ represents the largest integer not exceeding the real number $x)$. Taking the intersection of these two sets, denoted as $S_{n}$, ensures that any two numbers in $S_{n}$ have no divisibility or coprime relations. According to $n(\bmod 4)$, the number of elements in $S_{n}$ is:
$$
\begin{array}{l}
\left|S_{4 k}\right|=k,\left|S_{4 k+1}\right|=k, \\
\left|S_{4 k+2}\right|=k+1,\left|S_{4 k+3}\right|=k+1 .
\end{array}
$$

Use the pigeonhole principle to prove that the above $\left|S_{n}\right|$ is the maximum value of $|S|$. The number of target pigeonholes should be $\left|S_{n}\right|$. As long as the numbers in each pigeonhole have divisibility or coprime relations, it can ensure that $|S|$ does not exceed the number of pigeonholes.

First, set the pigeonholes as $\{4 t+1,4 t+2,4 t+3, 4 t+4\}\left(t=0, \cdots,\left[\frac{4}{n}\right]-1\right)$. It is found that $4 t+1$, $4 t+2$, and $4 t+3$ are pairwise coprime, and $4 t+2$ and $4 t+4$ are not coprime, and generally do not have a divisibility relation. Therefore, make an adjustment: take out each $2^{u}(2 t+1)(u \geqslant 2)$ and place it in the pigeonhole containing $4 t+2$.

The number of pigeonholes remains unchanged, but the size of the pigeonholes will differ. Each pigeonhole contains numbers of the form
$$
\left\{4 t+1,4 t+3,(2 t+1) \times 2^{u}\right\}
$$

Clearly, the numbers in each pigeonhole are pairwise coprime or have a divisibility relation, which meets the conditions of the problem.
The number of such constructed pigeonholes for
$$
n=4 k, 4 k+1,4 k+2,4 k+3
$$

are $k$, $k+1$, $k+1$, and $k+1$ respectively. Except for $n=4 k+1$, the number of pigeonholes is the same as the previously constructed $\left|S_{n}\right|$. Therefore, for $n=4 k+1$, the number of pigeonholes needs to be reduced by 1.
At this point, the last pigeonhole contains only the number $4 k+1$.
If 3 divides $(4 k+1)$, place $4 k+1$ in the pigeonhole $\left\{1,3,2^{t}\right\}$;

If 3 does not divide $(4 k+1)$, place $4 k+1$ in the pigeonhole $\{4 k-1, 4 k-2,4 k-3\}$.

It can be verified that the new pigeonholes still satisfy the condition that any two numbers have a divisibility or coprime relation, which meets the conditions of the problem.
Therefore, $|S|=\left|S_{n}\right|=\left[\frac{n+2}{4}\right]$.