朋友数必有相同

用抽屉原理。

每个人的朋友数只能是 $0,1,\dots,N-1$。

但不可能同时有人朋友数为 0（没人认识）且有人朋友数为 $N-1$（认识所有人），因为后者会认识前者。

因此实际可能的朋友数取值至多只有 $N-1$ 种，而有 $N$ 个人，于是至少两个人朋友数相同。

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Original Explanation

Since there are $N-1$ possibilities for friend-count and total $N$ people, atleast two people must have the same count.

Solution

The question can be solved by using the Pigeonhole Principle, which essentially states that if there are more pigeons than pigeonholes, then there must be at least one pigeonhole with more than one pigeon.

In this party of $N$ people, indexed from $1$ to $N$. The maximum number of friends a person can have is $N-1$, because one can be friends with every other person but not with oneself. So the possible number of friends one can have ranges from $0$ to $N-1$, a total of $N$ possibilities.

Note: If someone has zero friends, then someone else can have atmost $N-2$ friends, alternatively if someone has $N-1$ friends then someone else must have at-least $1$ friends.

Hence, the total possibilities of the count of friends ranges from $0$ to $N-2$ or from $1$ to $N-1$, i.e, total $N-1$ possibilities, but there are only $N$ people at the party.

Let's consider the count of friends as "pigeonholes" and people as "pigeons". Notice that there are $N$ people ("pigeons") and $N - 1$ different count of friends ("pigeonholes").

So, by the Pigeonhole Principle, at least two people must have the same count of friends in the party of $N$ people.

There's another way to solve this.

Proof by Contradiction: Assume that every person has a different number of friends from $0$ to $N - 1$. We would have a situation where one person has $0$ friends and another person has $N-1$ friends. This, however, contradicts the premise that friendships are symmetric, because if one person has $N-1$ friends, there cannot be someone with $0$ friends. Hence not everyone has a different number of friends