TW-TST 1998 Problem 6

In a group of $n\geq 4$ persons, every three who know each other have a common signal. Assume that these signals are not repeatad and that there are $m\geq 1$ signals in total. For any set of four persons in which there are three having a common signal, the fourth person has a common signal with at most one of them. Show that there three persons who have a common signal, such that the number of persons having no signal with anyone of them does not exceed $[n+3-\frac{18m}{n}]$ .