Re-interpreting in terms of graphs, $2$ people know each other if and only if there is an edge between their vertices in the corresponding graph $G$ and three people have a common signal if and only if they form a $K_3$/triangle in $G$. For a triangle $ABC$ in $G$, let $x(ABC)$ denote the number of vertices which are not contained in a triangle with any of $A,B,C$. Then $n-3-x(ABC)$ vertices form a triangle with at least one of $A,B,C$ (aside from $A,B,C$ themselves). Notice further that any fourth vertex $D$ can form a triangle with at most one of $A,B,C$ as otherwise the given condition is violated. Thus, among vertices contributing to $x(ABC)$, we can partition them into those adjacent to $A$, those adjacent to $B$, and those adjacent to $C$. Furthermore, any vertex contributing to $x(ABC)$ and adjacent to $A$ belongs to exactly one triangle with $A$, as otherwise the condition is again violated. We can argue similarly for $B$ and $C$. We now count pairs containing a triangle $ABC$ in $G$ and a vertex in $V(G) \setminus \{A,B,C\}$ incident to a triangle containing one of $A$, $B$, or $C$. As noted above, for a given triangle $ABC$, the number of pairs with the triangle being $ABC$ is $n-3-x(ABC)$. Letting $\mathcal{T}$ be the set of triangles in $G$, the count is then: \[\sum_{ABC \in \mathcal{T}} (n-3-x(ABC))\] As mentioned above, for a vertex $v$, the set of triangles containing $v$ are such that the pairs of the other two vertices are disjoint from each other. We can then get the count by looking at individual vertices. Letting the number of triangles incident to $v$ be $T(v)$, we choose the triangle around the vertex for the pair in $T(v)$ ways and we choose the vertex in the pair in $2(T(v)-1)$ ways. Then: \[\sum_{ABC \in \mathcal{T}} (n-3-x(ABC)) = \sum_{v \in V(G)} (2T(v)^2-2T(v))\] Notice that $\sum_{v \in V(G)} T(v)$ counts each triangle $3$ times, once for each vertex, so it is $3m$. Then: \[m(n-3) - \sum_{ABC \in \mathcal{T}} x(ABC) = \sum_{v \in V(G)} 2T(v)^2 - 6m\]\[\Longrightarrow m(n+3) - \sum_{ABC \in \mathcal{T}} x(ABC) = 2\sum_{v \in V(G)} T(v)^2\] Using Cauchy-Schwarz on the vectors $<T(v_1), \ldots, T(v_n)>$ and $<1, \ldots, 1>$, we get $\sum_{v \in V(G)} T(v)^2 \ge \frac{9m^2}{n}$. Rearranging the above: \[\sum_{ABC \in \mathcal{T}} x(ABC) \le m(n+3) - \frac{18m^2}{n}\] By the pigeonhole principle, some $ABC \in \mathcal{T}$ must be such that $x(ABC) \le \left\lfloor \frac{m(n+3)-\frac{18m^2}{n}}{m}\right\rfloor = \left\lfloor n+3 - \frac{18m}{n} \right\rfloor$.