People and acquaintances can be represented by a complete graph in two colours, where the vertices represent people, blue edges connect people who know each other, and red edges connect people who don't know each other. We call a set with $m$ vertices a blue (red) $m$-gon if all of the vertices are connected by a blue (red) edge. On the figures, red edges are represented by full lines, and blue edges by dashed lines.

Notice that the given condition implies that the graph contains a red or blue triangle (in the nontrivial cases where the graph has $n \ge 4$ vertices).

Without loss of generality, we can assume that the graph contains a red triangle. Let us observe the largest red $k$-gon. Let $\mathcal{A} = \{A_1, A_2, \dots, A_k\}$ be the set of its vertices. If the remaining $n-k$ vertices are all connected by blue edges, we can place the people from $\mathcal{A}$ in one room and everyone else in the other room, and the problem is solved.

We will show that the other case is not possible. Suppose there exist vertices $B, C \notin \mathcal{A}$ connected by a red edge. Then there exists $A_B \in \mathcal{A}$ which is connected to $B$ by a blue edge, because otherwise we would have a red $(k+1)$-gon. Analogously, there exists $A_C \in \mathcal{A}$ which is connected to $C$ by a blue edge.

We have two possibilities. If $A_B \neq A_C$, then it is clear (see the left figure above) that no three of the vertices $B, C, A_B, A_C$ can form a monochromatic triangle, which is in contradiction with the initial condition.

On the other hand, suppose $A_B = A_C = A$, for some $A \in \mathcal{A}$. Let $A' \in \mathcal{A}$ be an arbitrary vertex different from $A$. The right figure represents the situation between vertices $A, A', B$ and $C$. Clearly, in order to have a monochromatic triangle, $A'$ must be connected to $B$ and $C$ by a red edge. So, $B$ and $C$ are connected by a red edge, and they are both connected by a red edge with every vertex from $\mathcal{A} \setminus \{A\}$. But then $(\mathcal{A} \setminus \{A\}) \cup \{B, C\}$ is a red $(k+1)$-gon, which is again in contradiction with the maximality of $k$.