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, a formula can be derived:
$$
m_{\max }=\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right],
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
To facilitate the narrative, we first introduce the notation “+” and “-”.
For sets $A, B$, we denote
\[ A + B = A \cup B \]
\[ A + (-1)B = A - B = A \cap \complement B \]
Here, the “+” and “-” are different from those in real numbers and cannot be arbitrarily interchanged in the order of operations; they can only 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 \( M = N + \sum_{i=1}^{\infty}(-1)^{i} A_{p} \) meets the requirements.
Clearly, there exists \( k \in \mathbb{Z}_{+} \) such that
\[ p^{k-1} \leqslant n, \quad p^{2 l-1} \leqslant n \]
Consider \( B_{0}, B_{2}, \cdots, B_{2 l-2} \). It is easy to see that the elements in these sets do not repeat, and any element in them can be considered as a drawer. Clearly, 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 set \( N - \left(\bigcup_{i=0}^{l-1} C_{2 i}\right) \) and treat each element in it as a drawer.
Next, we 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 divisible by \( p^{2 i+1} \) but not by \( p^{2 i+2} \). Such numbers belong to \( A_{p^{2+1}} - A_{p^{2+2}} \),
and there are \( \left|A_{p^{2+1}}\right| - \left|A_{p^{2+2}}\right| = \left[\frac{n}{p^{2 i+1}}\right] - \left[\frac{n}{p^{2 i+2}}\right] \) such numbers.
Thus, the total number of drawers is
\[
\begin{array}{l}
\left|n - \left(\bigcup_{i=0}^{l-1} C_{2 i}\right)\right| + \sum_{i=0}^{l-1} \left|B_{2 i}\right| \\
= n - \sum_{i=0}^{l-1} 2 \left|B_{i}\right| + \sum_{i=0}^{l-1} \left|B_{2 i}\right| = n - \sum_{i=0}^{l-1} \left|B_{2 i}\right| \\
= n - \sum_{i=0}^{l-1} \left( \left[\frac{n}{p^{2+1}}\right] - \left[\frac{n}{p^{2+2}}\right] \right) = \sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right] .
\end{array}
\]

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