$4 \cdot 2117$ elves are sitting around a round table, and in front of each elf is a cup of milk. One elf pours all his milk evenly into the other cups. Then his first neighbor to the right does the same, followed by the next one, and so on. After the 7th elf has poured his milk into the other cups, each cup still contains as much milk as it did at the beginning, and the total amount of milk in all the cups is 3 liters. How much milk was there in each cup at the beginning?
［Solution］Initially, each cup contains milk of $\frac{6}{7}, \frac{5}{7}, \frac{4}{7}, \frac{3}{7}, \frac{2}{7}, \frac{1}{7}, 0$ liters, respectively.
In fact, after each elf pours milk to the others (giving each elf $\frac{1}{7}$ liter),

the original distribution can be obtained, just shifted by one position, and the total amount of milk $\frac{1+2+3+\cdots+6}{7}$ is exactly 3, which meets the condition of the problem.
The solution to this problem is unique.
In fact, we can assume that during the milk pouring process, when it is elf $A$'s turn to pour milk, $A$ has the most milk, $x$ liters. Since the milk pouring process can be repeated indefinitely, we can assume that $A$ pours milk first. After the 7 elves take turns pouring milk, since the other 6 elves pour no more than $\frac{x}{6}$ liters to $A$. However, the total amount of milk poured to $A$ is $x$ liters, so the other 6 elves each pour exactly $\frac{x}{6}$ liters to $A$. Therefore, according to the given conditions, each elf has $x$ liters of milk when it is their turn to pour, and after receiving $k$ portions, they have $\frac{k x}{6}$ liters $(k=1,2, \cdots, 6)$. Since the total amount is 3 liters, we have $\frac{x}{6}+\frac{2 x}{6}+\cdots+\frac{6 x}{6}=3$.
Thus, $x=\frac{6}{7}$.
Therefore, we obtain the solution mentioned above.
The uniqueness is proven.