We will prove the claim using mathematical induction on $n$. We will call the set $\{a_1, a_2, \dots, a_n\}$ good if it satisfies the given condition.

For $n=2$ the set $\{1,2\}$ is clearly good. Let us assume that for some positive integer $n \ge 2$ there exists a good set $\{a_1, a_2, \dots, a_n\}$. Let $a$ be the least common multiple of the set
$$\{a_1, a_2, \dots, a_n\} \cup \{a_j - a_i \mid 1 \le i < j \le n\}.$$
We claim that the set $\{a, a+a_1, a+a_2, ..., a+a_n\}$, containing $n+1$ elements, is also good.

Notice that for all $i \in \{1, 2, ..., n\}$ we have
$$\frac{(a+a_i)+a}{(a+a_i)-a} = \frac{2a+a_i}{a_i} = 2\frac{a}{a_i} + 1,$$
which is a positive integer since $a$ is divisible by $a_i$ (by definition).

Moreover, for all $1 \le i < j \le n$ we have
$$\frac{(a+a_j)+(a+a_i)}{(a+a_j)-(a+a_i)} = \frac{2a+(a_j+a_i)}{a_j-a_i} = 2\frac{a}{a_j-a_i} + \frac{a_j+a_i}{a_j-a_i},$$
which is a positive integer as well.

This completes the inductive step, thereby proving the claim.