Let $p_1, \dots, p_k$ be distinct prime numbers and for all $1 \le i \le n$ define $x_i$ to be $x_i = \prod_{j \in A_j} p_j$. We prove that these $x_i$'s satisfy the problem's conditions.

$$1 \le i, j \le k \implies A_i \cap A_j \ne \emptyset \implies \exists l : l \in A_i, l \in A_j \implies p_i p_j \mid x_l \\ l \in A_i \implies p_j \mid x_l \mid \operatorname{lcm}_{m \in A_i}\{x_m\}$$
So for every $1 \le i, j \le k$, $p_j \mid x_l \mid \operatorname{lcm}_{m \in A_i}\{x_m\}$, which results in $x_l \mid \operatorname{lcm}_{m \in A_i}\{x_m\} = p_1 \dots p_k$.

On the other hand,
$$p_i \mid \operatorname{lcm}_{m \notin A_i}\{x_m\} \iff \exists j : p_i \mid x_j,\ j \notin A_i \iff j \in A_i,\ j \notin A_i$$
which clearly is a contradiction, therefore
$$\operatorname{lcm}_{m \notin A_i}\{x_m\} \le \frac{p_1 \dots p_k}{p_i} < p_1 \dots p_k = \operatorname{lcm}_{m \in A_i}\{x_m\}$$
which yields our result.