5. On the island, there are 217 residents, 17 of whom are knights, and the remaining 200 are tricksters. One day, a private detective arrived on the island and decided to find out who is who. For this, he asked each resident to write a list of 200 people they consider to be tricksters. The detective does not know who wrote which list, but he knows that the knights honestly wrote down 200 real tricksters, while the tricksters could have written down any residents of the island (including themselves). Can the detective always identify at least 13 residents who are definitely tricksters based on the received lists? All residents of the island have different names, so a name written in the list always uniquely corresponds to one resident of the island.
Answer: Yes, it can always.

Solution. We will combine identical lists into groups. Let's call a group important if it contains at least 17 lists. Note that the number of important groups is no more than 12 (otherwise, the number of residents is at least $13 * 17 = 221$, which is more than 217).

Since all the knights should have ended up with the same list (all wrote down the same tricksters), the lists of all the knights ended up in one group.

Since there are 17 knights, this group contains at least 17 lists, meaning it is important. Thus, in one of the important groups, all the tricksters are listed.

We will prove that there are at least 13 residents who are listed in all important groups. We will mark the people who are not listed in the important groups. In each group, only 17 residents of the island are not listed. Therefore, since there are no more than 12 important groups, a total of no more than 17 * 12 = 204 residents will be marked. At least 13 residents will not be marked, meaning they are listed in all important groups. Regardless of which important group contains the lists of the knights, these 13 residents are definitely tricksters.

Thus, to detect the tricksters, the private detective should combine identical lists into groups and find 13 names that appear in all important (containing at least 17 lists) groups.

Criteria. Only the answer - 0 points. Considering only individual special cases (for example, that all residents wrote down the real tricksters) - 0 points. The solution contains the idea of combining identical lists into groups and highlighting groups with a size of 17 or more lists (but no further progress) - 1 point.