China Girls' Mathematical Olympiad

For every $i$ with $1 \le i \le n$, we define set
$$S_i = \{b \mid b \text{ is an element in } S \text{ and is divisible by } a_i\}.$$
There are $\lfloor \frac{m}{a_i} \rfloor$ elements in $S_i$. Since every element in $S$ is not divisible by any two distinct elements in $T$, it follows that $S_i \cap S_j = \emptyset$ for $1 \le i < j \le n$. Thus
$$\sum_{i=1}^{n} \lfloor \frac{m}{a_i} \rfloor = \sum_{i=1}^{n} |S_i| \le |S| = m.$$
Note that $\frac{m}{a_i} < \lfloor \frac{m}{a_i} \rfloor + 1$. It follows that
$$\sum_{i=1}^{n} \frac{m}{a_i} < \sum_{i=1}^{n} (\lfloor \frac{m}{a_i} \rfloor + 1) = \sum_{i=1}^{n} \lfloor \frac{m}{a_i} \rfloor + \sum_{i=1}^{n} 1 \le m + n.$$
So the desired result is obtained.