 Solution

Let $N$ count everyone present, including the host. The grouping, seating and departure conditions give the modular congruencesThe first two imply $N=39+45k$, and oddness forces $k$ even. By the Chinese remainder theorem, the complete solution isBecause not every invitee attended, at most $y-1$ guests came, so $N\le y$. For $y=120$, the only positive possibility is 39 people. It is feasible: seven groups of five and one of four; three tables of ten and one of nine; nineteen departing pairs and the host.

For $y=150$, both 39 and 129 people are possible. The latter uses twenty-five groups of five and one of four, three tables of ten and eleven of nine, and sixty-four departing pairs. Thus the larger invitation count destroys uniqueness; it does not force a different attendance.