Pick $k$ distinct primes $p_1, \dotsc, p_k$. Let $x_i$ be the product of all $p_j$ where $i\in A_j$. By assumption, for any $j, m$, some element of $A_j$ is also in $A_m$, so \[\mathrm{lcm}_{i\in A_j}\{x_i\}=p_1\dotsm p_k\]because every prime is represented at least once. On the other hand, $A_j^{c}$ has no element in common with $A_j$, so \[\mathrm{lcm}_{i\not\in A_j}\{x_i\}\leq \frac{p_1\dotsm p_k}{p_j}.\]Hence this works.

Let's induct on $k$. The case $k=1$ is trivial; consider the induction step. Suppose we've chosen $x_i$'s so that the LCM condition holds for $A_1,\cdots ,A_k$. Add a new subset $A_{k+1}$. Choose a huge prime $p$. Define new numbers $x'_i$s a as follows: $x'_i=px_i$ if $i\in A_{k+1},$ and $x'_i=x_i$ otherwise. We claim that these work. For $A_i$ with $1\le i\le k$, the LHS of the LCM inequality is now $p$ times as before (since $A_i$ necessarily has some element common with $A_{k+1}$), and the RHS is either unchanged or $p$ times as before, so we're good. It remains to ensure $\operatorname{lcm}_{i\in A_{k+1}}\{x'_i\}>\operatorname{lcm}_{i\not\in A_{k+1}}\{x'_i\}$ i.e. $ p\operatorname{lcm}_{i\in A_{k+1}}\{x_i\}>\operatorname{lcm}_{i\not\in A_{k+1}}\{x_i\}$, but that can be fixed by taking a large enough $p$, so we're done. $\blacksquare$

A bit easy for Iran: just induct on $k$ and for $A_{k+1}$ just multiply all elements in $\{x_i\in A_{k+1}\}$ by a sufficiently large prime.

a1267ab wrote: Pick $k$ distinct primes $p_1, \dotsc, p_k$. Let $x_i$ be the product of all $p_j$ where $i\in A_j$. By assumption, for any $j, m$, some element of $A_j$ is also in $A_m$, so \[\mathrm{lcm}_{i\in A_j}\{x_i\}=p_1\dotsm p_k\]because every prime is represented at least once. On the other hand, $A_j^{c}$ has no element in common with $A_j$, so \[\mathrm{lcm}_{i\not\in A_j}\{x_i\}\leq \frac{p_1\dotsm p_k}{p_j}.\]Hence this works. One of the most elegant solutions I have ever seen Where did you get the motivation for this solution please tell me