Let $n>2$ be a given positive integer. There are $n$ guests at Georg's bachelor party and each guest is friends with at least one other guest. Georg organizes a party game among the guests. Each guest receives a jug of water such that there are no two guests with the same amount of water in their jugs. All guests now proceed simultaneously as follows. Every guest takes one cup for each of his friends at the party and distributes all the water from his jug evenly in the cups. He then passes a cup to each of his friends. Each guest having received a cup of water from each of his friends pours the water he has received into his jug. What is the smallest possible number of guests that do not have the same amount of water as they started with?
Answer: 2.

If there are guests $1,2, \ldots, n$ and guest $i$ is friends with guest $i-1$ and $i+1$ modulo $n$ (e.g. guest 1 and guest $n$ are friends). Then if guest $i$ has $i$ amount of water in their jug at the start of the game, then only guest 1 and $n$ end up with a different amount of water than they started with.

To show that there always will be at least two guests with a different amount of water at the end of the game than they started with, let $x_{i}$ and $d_{i}$ be the amount of water and number of friends, respectively, that guest $i$ has. Define $z_{v}=x_{v} / d_{v}$ and assume without loss of generality that the friendship graph of the party is connected. Since every friend has at least one friend, there must exist two guests $a$ and $b$ at the party with the same number of friends by the pigeonhole principle. They must satisfy $z_{a} \neq z_{b}$. Thus, the sets

$$
S=\left\{c \mid z_{c}=\min _{d} z_{d}\right\} \text { and } T=\left\{c \mid z_{c}=\max _{d} z_{d}\right\}
$$

are non-empty and disjoint. Since we assumed the friendship graph to be connected, there exists a guest $c \in S$ that has a friend $d$ not in $S$. Let $F$ be the friends of $c$ at the party. Then the amount of water in $c$'s cup at the end of the game is

$$
\sum_{f \in F} z_{f} \geqslant z_{d}+\left(d_{c}-1\right) z_{c}>d_{c} \cdot z_{c}=x_{c}
$$

Thus, $c$ ends up with a different amount of water at the end of the game. Similarly, there is a guest in $T$ that ends up with a different amount of water at the end of the game than what they started with.