3. Once, 20 people gathered in a castle, 11 of them were sages who always tell the truth, and the other 9 were jesters who always lie. Around a table, $n \geq 3$ people sat, each of whom claimed, "Exactly one of my two neighbors is a jester." For which $n$ is this possible? (It is necessary to find all possible $n$ and explain why others cannot be).
Answer. $n=3,4,5,6,7,8,9,12,15$.

Solution. First, there is a case where all the people at the table are jesters: in this case, all $n \leq 9$ are suitable.

Now, let there be at least one sage at the table. Let this sage be $A$, and his neighbors be sage $B$ and jester $C$. Then, on the other side of $B$ sits a jester, and on the other side of $C$ sits a sage. Continuing this reasoning, we get that the people sitting at the table are divided into non-overlapping triples, each containing two sages and one jester. Therefore, $n=3k$, and at the table there are $k$ jesters and $2k$ sages. From the condition $k \leq 9$ and $2k \leq 11$, it follows that $k \leq 5$. Thus, two more possible values are added: $n=12, n=15$.

Criteria. Full solution - 7 points. For losing one case, deduct 2 points. If 3 or more cases are lost - 0 points. All variants are found, but there is no proof that there can be no others - 3 points.