For the set $N=\{1,2, \cdots, n\}$, select $m$ numbers such that none of them is $p$ times another. Find the maximum value of $m$.


For this problem, we can obtain a formula:
$$
m_{\max }=\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right] \text {, }
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Proof: For convenience, we introduce the notations “+” and “-”.
For sets $A$ and $B$, we denote
$A+B=A \cup B$,
$A+(-1) B=A-B=A \cap \overline{B}$.
Here, the “+” and “-” are different from those in real numbers and cannot be arbitrarily interchanged; they must be computed from left to right. Additionally,
if $A \cap B=\varnothing$, then $|A+B|=|A|+|B|$;
if $A \supseteq B$, then $|A-B|=|A|-|B|$.
(1) First, we prove: $m \geqslant \sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right]$.
Let $A_{i}=\{u \mid u=k i, 1 \leqslant u \leqslant n, k \in \mathbb{N}\}\left(i \in \mathbb{Z}_{+}\right)$.
It is easy to see that $\left|A_{i}\right|=\left[\frac{n}{i}\right]$.
Thus, $N \supseteq A_{p} \supseteq A_{p^{2}} \supseteq A_{p^{3}} \supseteq \cdots$.
We can prove that the set $M=N+\sum_{i=1}^{\infty}(-1)^{i} A_{p^{i}}$ meets the requirements.
Clearly, there exists $k \in \mathbb{Z}_{+}$ such that
$p^{k-1} \leqslant n, p^{2 l-1} \leqslant n$.

Consider $B_{0}, B_{2}, \cdots, B_{2 l-2}$. It is easy to see that the numbers contained in these sets are not repeated, and any element in them can be considered as a drawer, and each drawer can only take one number.

Let $C_{i}=\bigcup_{A \in B_{i}} A$, i.e., $C_{i}$ is the union of all elements in $B_{i}$. Consider the elements in the set $N-\left(\bigcup_{i=0}^{l-1} C_{2 i}\right)$ as drawers.
We now prove that the total number of drawers is $\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right]$. To do this, we compute $\left|B_{2 i}\right|$.
Clearly, any element in $B_{2 i}$ consists of two numbers, where the largest number in $N$ can only be divided by $p^{2 i+1}$ but not by $p^{2 i+2}$, such numbers belong to $A_{p^{2 i+1}}-A_{p^{2 i+2}}$,
and there are $\left|A_{p^{2 i+1}}\right|-\left|A_{p^{2 i+2}}\right|=\left[\frac{n}{p^{2 i+1}}\right]-\left[\frac{n}{p^{2 i+2}}\right]$ such numbers.
Thus, the total number of drawers is
$$
\begin{array}{l}
\left|n-\left(\bigcup_{i=0}^{l-1} C_{2 i}\right)\right|+\sum_{i=0}^{l-1}\left|B_{2 i}\right| \\
=n-\sum_{i=0}^{l-1} 2\left|B_{i}\right|+\sum_{i=0}^{l-1}\left|B_{2 i}\right|=n-\sum_{i=0}^{l-1}\left|B_{2 i}\right| \\
=n-\sum_{i=0}^{l-1}\left(\left[\frac{n}{p^{2 i+1}}\right]-\left[\frac{n}{p^{2 i+2}}\right]\right)=\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right] .
\end{array}
$$

By the pigeonhole principle, each drawer can take at most one number.
Therefore, $m \leqslant \sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right]$.
In summary, $m_{\max }=\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right]$.