# Solution.

The point is that after the first gnome pours the milk (giving $\frac{1}{7}$ to each of the others), the distribution remains exactly the same but shifted by one gnome, and the sum $(1+2+\ldots+6): 7$ is exactly 3. It remains to prove that there are no other solutions. Let $x$ be the maximum amount of milk that any gnome $\Gamma$ has at any point during the pouring. Then, after another cycle of seven "pourings" (which can be continued indefinitely, so $\Gamma$ can be considered the first in the cycle), $\Gamma$ will have accumulated no more than $6 \cdot \frac{x}{6}=x$ liters of milk; and equality is possible only if each of the six other gnomes pours exactly $\frac{x}{6}$ liters of milk into $\Gamma$'s cup. Thus, from the condition, it follows that each gnome pours the same amount $x$ of milk and after receiving $k$ portions, he has $\frac{k x}{6}$ liters in his cup ($k=1$, 2, ..., 6). Knowing from the condition that $(x+2 x+\ldots+6 x): 6=\frac{7}{2} x=3$, we find the unique solution.

Answer. 6/7, 5/7, 4/7, 3/7, 2/7, 1/7, 0 liters.
2. (7 points) Seven dwarfs are sitting around a round table. In front of each of them is a mug. Some of these mugs contain milk. One of the dwarfs pours all his milk equally into the mugs of the others. Then his right neighbor does the same. Then the next right neighbor does the same, and so on. After the last, seventh dwarf has poured all his milk to the others, each mug contains the same amount of milk as it did initially. Together, all the mugs contain 3 liters of milk. How much milk was initially in each mug?