I thought I would never post on aops again but I really like this problem and still no one has posted a solution Let's see the problem as a graph $G$ with edges of two colors, red if $A$ know $B$ and blue in the other case Let $v_1, v_2, ..., v_n$ the vertices of $G$ Lema: Let $Q$ a sub-graph of $G$. Then, there exist a vertex on $Q$ wich have all their edges (in $Q$) of the same color Proof: Let's apply induction on the number of vertices in $Q$, the base case is trivial Let's asumme the Lema for every $Q$ with at most $k$ vertices ...($i$) Suppose it doesn't works for any sub-graph $P$ with $k+1$ vertices, so each of their vertices has at least 1 red edge and 1 blue edge ...($ii$) WLOG $P$ is $v_1, v_2, ..., v_{k+1}$ , by ($i$), in $v_1, v_2, ..., v_k$ there exist a vertex which have all their edges of the same color, WLOG that vertex is $v_1$ and their edges to $v_2, v_3, ..., v_k$ are blue By ($ii$): the edge of $v_1$ and $v_{k+1}$ is red. Also, by ($ii$): there exist a vertex in $v_2, ..., v_k$ which is connected to $v_{k+1}$ with a blue edge, WLOG is $v_2$ By ($ii$): there exist a vertex in $v_3, ..., v_k$ which is connected to $v_2$ with a red edge, WLOG is $v_3$ Then, if we seat $v_2, v_3, v_1, v_{k+1}$ around a table in that order, the condition of the problem doesn't works, which is a contradiction So, the Lema works Then, let's define the secuence of vertices of G: $w_1, w_2, ..., w_n$ If we have $w_1, w_2, ..., w_x$ , then, by the lema, in the rest of vertices there exist a vertex with all their edges of the same color, that vertex will be $w_{x+1}$ . If the edges of $w_{x+1}$ are red, we will call it fyre tipe, if the edges are blue, we will call it water type Finally, we divide the graph in one group of the water type vertices and one group of the fire type vertices and it will satisfy what we are looking for

Take the obvious graph interpretation. A bit of casework translates the condition to every four students having some student adjacent to either everyone else or nobody else. Take the largest complete subgraph $K$. If at most $1$ vertex isn't in it, we're done. Otherwise, pick two arbitrary vertices $v,w$ not in $K$ and suppose that they're adjacent. By maximality we can find two vertices $x,y$ in $K$, the first of which isn't adjacent to $v$, and the second of which isn't adjacent to $w$. If these vertices are distinct then we're done by considering $\{v,w,x,y\}$, since $\overline{xy}$ is an edge. Otherwise, $v$ and $w$ are both adjacent to every vertex in $K$ other than $x=y$, so $(K \setminus \{x\}) \cup \{v,w\}$ is a larger complete subgraph. $\blacksquare$