Three. (50 points) We call non-empty sets $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if:
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j}=\varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 13-partition $A_{1}, A_{2}, \cdots, A_{13}$ of $A=\{1,2, \cdots, m \}$, there must exist a set $A_{i}(1 \leqslant i \leqslant 13)$, in which there are two elements $a, b$ satisfying
$$
b<a \leqslant \frac{9}{8} b \text {. }
$$
(1) First, prove that $m \geqslant 117$.
If not, $m1+\frac{13}{104} \geqslant \frac{9}{8}$.
This contradicts $b<a \leqslant \frac{9}{8} b$. Therefore, $m \geqslant 117$.
(2) Next, prove that $m=117$ satisfies the condition.

Because, in this case, the largest 14 numbers $104,105, \cdots, 117$ are distributed among 13 sets. By the pigeonhole principle, there must be two numbers $a$ and $b (b < a)$ in the same set $A_{i}(1 \leqslant i \leqslant 13)$, then
$$
104 \leqslant b<a \leqslant 117=\frac{9}{8} \times 104 \leqslant \frac{9}{8} b .
$$

In summary, the smallest $m=117$.