Let points $A_1, A_2, \cdots, A_{100}$ represent 100 guests. We connect two points using a red segment if they know each other, and using a blue segment if they don't know each other. Hence, from every point, there are at most 32 blue segments. It follows that there at least three points on the other ends of these blue segments are connected with each other in red. Let them be $A_1, A_2,$ and $A_3$. From these three points, there are most 96 blue segments. If we remove all these 96 segments together with the other ends, there must left at least one point, let it be $A_4$. Then $A_1, A_2, A_3,$ and $A_4$ are connected in red.