Prove that at a party with at least two people, there are two people who know the same number of people...

Let $n$ be the number of party-goers. The maximum number of people a person can know is $n-1$ and the minimum number he/she can know is 1 (by assumption), giving us $n-1$ possibilities for the number of people someone can know. Every single person must be assigned one of these $n-1$ possible numbers but since there are $n$ party-goers one of these numbers must be used twice due to the pigeonhole principle i.e. two party-goers know the same number of people.

There are two cases to consider:

1. Assume there is a someone at the party, lets say Joe, who knows everyone else at the party. He must know $n-1$ people. In this case, everybody else at the party must know at least know Joe, and the minimum number of people a person can know is $1$. This gives us the set $\{1, 2, ... n - 1\}$ which represents the possible number of people each person can know.
2. Assume there is a party crasher, Harry, who doesn't actually know anybody. This means that even a socialite like Joe can't possibly know everyone, so the maximum number of people a person can know is $n - 2$. This gives us the set $\{0, 1, ..., n - 2\}$.

Both these sets have n-1 elements. Since there are $n$ people at the party, and $n-1$ possibilities for the number of people each person can know, it follows that there must be at least two people who know the same number of people.