Given positive integers $m, a, b, (a, b)=1$. $A$ is a non-empty subset of the set of positive integers, such that for any positive integer $n$, either $a n \in A$ or $b n \in A$. For all sets $A$ satisfying the above property, find the minimum value of $|A \cap \{1, 2, \cdots, m\}|$.

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The problem is to find the minimum size of the intersection of $A$ with the set $\{1, 2, \cdots, m\}$ for all sets $A$ that satisfy the given conditions.
(1) When $a=b=1$, we have $A \cap\{1,2, \cdots, m\}=\{1,2, \cdots, m\}$, $|A \cap\{1,2, \cdots, m\}|=m$.
(2) Without loss of generality, assume $a>b$. Let
$A_{1}=\left\{k \mid\right.$ if $a^{\alpha} \mid k, a^{\alpha+1} \nmid k$, then $\alpha$ is odd $\}$.
Now verify that $A_{1}$ satisfies the condition.
For any positive integer $n$, let $n=a^{\alpha} n_{1}, a \nmid n_{1}$.
If $2 \mid \alpha$, then $a n=a^{\alpha+1} n_{1} \in A_{1}$;
If $2 \nmid \alpha$, then $b n=a^{\alpha} b n_{1} \in A_{1}$.
Thus, $\left|A_{1} \cap\{1,2, \cdots, m\}\right|=\sum_{i=1}^{\infty}(-1)^{i+1}\left[\frac{m}{a^{i}}\right]$.
(3) For any positive integer $n$, take $c_{n}=a$ or $b$, such that $n c_{n} \in A$. Let $B=\left\{c_{1}, 2 c_{2}, 3 c_{3}, \cdots\right\}$. Therefore,
$|A \cap\{1,2, \cdots, m\}| \geqslant|B \cap\{1,2, \cdots, m\}|$.
For any $n$, let $n=a^{\alpha} n_{1}, a \nmid n_{1}$, take
$d_{n}=\left\{\begin{array}{ll}a, & \text { if } 2 \mid \alpha ; \\ b, & \text { if } 2 \nmid \alpha .\end{array}\right.$
Let $B_{n}=\left\{d_{1}, 2 d_{2}, \cdots, n d_{n},(n+1) c_{n+1},(n+2) c_{n+2}, \cdots\right\}$,
$B_{0}=B$.
We now prove:
$\left|B_{i} \cap\{1,2, \cdots, m\}\right| \geqslant\left|B_{i+1} \cap\{1,2, \cdots, m\}\right|$,
$i=0,1, \cdots$.
Use mathematical induction on $i$.
(i) When $i=0$, if $c_{1}=b$, then $c_{1}i+1$ when, $(i+1) c_{i+1}=(i+1) bm$, then
$(i+1) d_{i+1}=(i+1) a$
$>(i+1) b=(i+1) c_{i+1}>m$.
In this case, the equality in (2) holds.
ii) If $2 \nmid a$, then $d_{i+1}=b$. Thus, $c_{i+1}=a$. In this case,
$(i+1) d_{i+1}=a^{\alpha} b n_{i}=i^{\prime} d_{i^{\prime}}, i^{\prime}=a^{\alpha-1} b n_{i}<i+1$.
Hence, $|B_{i} \cap\{1,2, \cdots, m\}|$
$$
\begin{aligned}
= & \left|\left\{d_{1}, 2 d_{2}, \cdots, i d_{i},(i+1) c_{i+1}, \cdots\right\} \cap\{1,2, \cdots, m\}\right| \\
\geqslant & \left|\left\{d_{1}, 2 d_{2}, \cdots, i d_{i},(i+2) c_{i+2}, \cdots\right\} \cap\{1,2, \cdots, m\}\right| \\
= & \mid\left\{d_{1}, 2 d_{2}, \cdots, i d_{i},(i+1) d_{i+1},(i+2) c_{i+2}, \cdots\right\} \cap \\
& \{1,2, \cdots, m\} \mid \\
= & \left|B_{i+1} \cap\{1,2, \cdots, m\}\right| .
\end{aligned}
$$

Therefore, (2) holds.
This proves that (2) holds for all $i$.
Let $n$ be the largest positive integer such that $n d_{n} \leqslant m$. Then by (1) and (2),
$$
\begin{array}{l}
|A \cap\{1,2, \cdots, m\}| \\
\geqslant|B \cap\{1,2, \cdots, m\}|=\left|B_{0} \cap\{1,2, \cdots, m\}\right| \\
\geqslant\left|B_{n} \cap\{1,2, \cdots, m\}\right|=\left|A_{1} \cap\{1,2, \cdots, m\}\right| \\
=\sum_{i=1}^{\infty}(-1)^{i+1}\left[\frac{m}{a^{i}}\right] .
\end{array}
$$