Bulgarian Mathematical Olympiad

Solution:

Consider a group with maximal number of people such that any two of them are not familiar. It is clear that if there are $l$ people in this group, then $l \leq m-1$. Moreover, the maximality of $l$ implies that any of the other $N-l$ people is familiar to at least one of the $l$ people in the group. Hence some of these $l$ people is familiar to at least $\frac{N-l}{l}$ people. Since
$$\frac{N-l}{l} = \frac{N}{l} - 1 \geq \frac{N}{m-1} - 1 = n + \frac{1}{m-1} - 1 > n-1$$
there is a person who is familiar to $n$ people.

Let $N = m n - n = (m-1) n$ and consider $m-1$ groups by $n$ people such that any two people from one group are familiar and there are no familiar people from different groups. Then among any $m$ people there are two from one and the same group, i.e. they are familiar. On the other hand, any of the people is familiar to $n-1$ of the other and hence the statement is not true if $N < m n - n + 1$.