In a mathematical competition, in which 6 problems were
posed to the participants, every two of these problems were solved by more
than 2/5 of the contestants.

Moreover, no contestant solved all the 6 problems.

Show that there are at least 2 contestants who solved exactly 5 problems
each.

source: IMO 2005

Oh, I see where I went wrong.

If there are 7 contestants, each of the 15 pairs needs to be solved by 3 of them, for a minimum total of 45 solved solved pairs. If each student solves 4 problems, then there are only 7*6 = 42 solved pairs, so at least one student must solve 5 problems to bring the total up to 46 solved pairs.

I am not yet clear why 2 contestants must solve 5, but I guess it is because of overlaps. I'll pursue it further this evening.