Example 15 Let $I=\{1,2,3, \cdots, n\}$. Find the maximum positive integer $m$, such that there exists an $m$-element subset $S$ of $I$, satisfying:
(1) No number in $S$ is a multiple of another number in $S$;
Let $S_{0}=\left\{2 k \mid k \in \mathbf{N}_{+},\left[\frac{n+2}{4}\right] \leqslant k \leqslant 2\left[\frac{n+2}{4}\right]-1\right\}$. Since $2\left(2\left[\frac{n+2}{4}\right]-1\right) \leqslant 4 \times \frac{n+2}{4}-2=n$, it follows that $S_{0} \subseteq I, |S_{0}|=\left[\frac{n+2}{4}\right]$, and $S_{0}$ satisfies conditions (1) and (2). Therefore, the maximum positive integer $m \geqslant\left[\frac{n+2}{4}\right]$.

Next, let $S \subseteq I$ be any $m$-element set that satisfies conditions (1) and (2). For any $a \in S$, there exists a unique non-negative integer $k_{0}$ such that $2^{k_{0}} a \in S_{1}=\left\{\left[\frac{n}{2}\right]+1,\left[\frac{n}{2}\right]+2, \cdots, n\right\}$. We define a mapping $f$ from $S$ to $S_{1}$ by letting $a$ correspond to $2^{k_{0}} a$. We now prove that $f$ is injective.

Indeed, for any $a, b \in S$ with $a \neq b$, if $f(a) = f(b)$, then $2^{k_{a}} a = 2^{k_{b}} b$. This implies that $a$ divides $b$, which contradicts the fact that $S$ satisfies condition (1). Therefore, $f(a) \neq f(b)$, meaning $f$ is injective.

Furthermore, for any $a, b \in S$ with $a \neq b$, by condition (2), we have $(a, b) > 1$. Hence, $|f(a), f(b)| = \left(2^{k_{a}} a, 2^{k_{b}} b\right) > 1$. However, any two consecutive numbers in $S_{1}$ are coprime, and when $n$ is odd, $n-2, n-1, n$ are pairwise coprime. Therefore, the image set $f(S)$ does not contain any two consecutive integers in $S_{1}$, and when $n$ is odd, $f(S)$ contains at most one of $n-2, n-1, n$. Thus,
$$
m = |S| = |f(S)| \leqslant \left[\frac{n+2}{4}\right].
$$

In conclusion, the maximum value of $m$ is $\left[\frac{n+2}{4}\right]$.