Example 1 Given positive integers $r$ and $n$, find the smallest positive integer $m$, satisfying: if the set $S=\{1,2, \cdots, m\}$ is arbitrarily divided into $r$ pairwise disjoint subsets $A_{1}, A_{2}, \cdots, A_{r}$, then there must exist two positive integers $a, b$ belonging to the same subset $A_{i}(1 \leqslant i \leqslant r)$ and $b<a \leqslant \frac{n+1}{n} b$.
First, partition $S=\{1,2, \cdots, m\}$ into $r$ pairwise disjoint subsets, $A_{1}$, $A_{2}, \cdots, A_{r}$, where $A_{i}=\{k \mid k \in S$ and $k \equiv i(\bmod r)\}(i=1,2, \cdots, r)$.
Thus, for each $A_{i}(1 \leqslant i \leqslant r)$ and any $a, b \in A_{i}(a>b)$, we have $a-b \geqslant r$, and $b \leqslant a-r \leqslant m-$ $r$. Therefore,
$$
\frac{a}{b}=1+\frac{a-b}{b} \geqslant 1+\frac{r}{m-r} .
$$

Thus, when $m-r \geqslant n r$, i.e., $m \geqslant (n+1) r$, for any $a, b \in A_{i}(a>b)$, we have $\frac{a}{b} \geqslant 1+\frac{r}{m-r} \geqslant 1+\frac{r}{n r} = 1+\frac{1}{n}$, which means $a>\frac{n+1}{n} b$.
Hence, when $m \geqslant (n+1) r$, for any $a, b \in A_{i}(a>b)$, if $n r \leqslant b<a \leqslant n r+r$, then $a-b \leqslant r$, $b \geqslant n r$, and thus $1<\frac{a}{b}=1+\frac{a-b}{b} \leqslant 1+\frac{r}{n r}=\frac{n+1}{n}$, i.e., $b<a \leqslant \frac{n+1}{n} b$. Therefore, when $m \geqslant (n+1) r$, the condition of the problem is satisfied.
In summary, the minimum value of $m$ is $m_{0}=(n+1) r$.