If at least one of these $n$ people know none of the rest people, then among the remaining $(n-1)$ guests, one at most know $(n-2)$ people. Hence, overall, the number of other guests one known is from $0$ to $(n-2)$.

If every one knows at least one other guest, then the number of other guests one can know is from $1$ to $(n-1)$.

Either way, there are $(n-1)$ possibilities. It follows that there are at least two of them knows the same number of other guests, by the pigeonhole principle.