Seven dwarfs sat around a round table, each with a bowl in front of them, containing a total of half a liter of milk. (Some of the bowls might have been empty.) One of the dwarfs stood up and distributed the milk in his bowl equally among the others. Then, the dwarf sitting to his right did the same, and so on. After the seventh dwarf had also distributed his milk, everyone had exactly the same amount of milk in their bowls as they had before the distribution. How much milk was originally in each of the bowls?
Let $x_{i}$ denote the amount of milk that the $i$-th dwarf spilled when it was his turn. Suppose that among these, $x_{k}$ is the (one of the) largest. Since after one round the amount of milk in the pail was the same as at the beginning, the refilling will continue, and the $k$-th dwarf will spill the same amount as before. Let's examine how the amount of milk in the pail changed. First, it was emptied, then $\frac{x_{k+1}}{6}$ liters were added, followed by $\frac{x_{k+2}}{6}$ liters, and so on, a total of six times. Since $x_{i} \leq x_{k}$ for all $i$, after a full round, at most $6 \frac{x_{k}}{6}=x_{k}$ liters could be in the pail; on the other hand, we know that exactly $x_{k}$ liters were in it, which can only be if $x_{l}=x_{2}=\ldots=x_{7}$. Let this common value be $x$.

Now let's examine the spillings mentioned in the problem again! The first dwarf in the sequence had $x$ liters of milk in the pail. After the second dwarf spilled his milk, the (then empty) pail of the first dwarf received $\frac{x}{6}$ liters of milk five more times, so at the end of the distribution, and thus at the beginning of the next, he had $\frac{5 x}{6}$ liters of milk. Similarly, the third dwarf had $\frac{4 x}{6}$ liters, and then in sequence $\frac{3 x}{6}, \frac{2 x}{6}, \frac{x}{6}, \frac{0 x}{6}$, so that's how much he had at the beginning. According to the conditions,

$$
\frac{6 x}{6}+\frac{5 x}{6}+\ldots+\frac{x}{6}+0=\frac{1}{2}, \quad \text { from which } \quad x=\frac{6}{42}
$$

Therefore, at the beginning, the pail contained $\frac{6}{42}, \frac{5}{42}, \frac{4}{42}, \frac{3}{42}, \frac{2}{42}, \frac{1}{42}, 0$ liters of milk, respectively.