 Solution

Discard ground points lying in no set. For every point $x$ that lies in at least two members, formEvery pair of indices lies in exactly one $L_x$, and no $L_x$ equals $[n]$ because the total intersection is empty. Thus the $L_x$ form a finite linear space on the $n$ indices. The number of its lines through index $i$ is at most $a_i$, with equality unless $A_i$ contains private points.

The De Bruijn--Erdos pair-covering inequality says that if $r_i$ is the number of lines through point $i$ in a nontrivial finite linear space on $n$ points, thenApplying it here givesMoreover, a pair of ground points can lie in at most one $A_i$, since two different members meet in only one point. HenceCombining the inequalities gives $\left(\genfrac{}{}{0ex}{}{m}{2}\right)\ge \left(\genfrac{}{}{0ex}{}{n}{2}\right)$, and therefore $m\ge n$.