Let $n = 16$, and suppose for contradiction that $5$ meetings suffice.
Construct a bipartite graph $S ∪ T$ with $S$ the set of $5$ $K_ns$ and $T$ the vertices of the $K_{2n}$ ; draw $n$ red edges from fixed $s ∈ S$ to the $n$ vertices $t ∈ T$ such that $t ∈ s$; draw blue edges between any two vertices in $T$ sharing a common $K_n$. We need every two vertices of $T$ to have a blue edge. There are $5n$ red edges, so some $t ∈ T$ has red-degree less than $5n/2n$, so at most $2$, say to $s_1$, $s_2$.

But every $t≠t$ in $T$ shares a blue edge with $t$, so $N(s_1) ∪ N(s_2) = T$. But $deg$ $s_1 =$ $deg$ $s_2 = n$, $|T| = 2n$, and $t ∈ N(s_1) ∩ N(s_2)$, so we get a contradiction by PIE. Note that the same logic will not work when we change $K_n$ and $K_{2n}$ to $K_n$ and $K_{mn}$ for some $m > 2$.

For the construction, partition the $32$ members into $4$ groups of $8$, and let $G_i$ ∪ $G_j$ be the six meetings for $1 ≤ i < j ≤ 4$.