6. The participants in a conference of scoundrels (who always lie) and knights (who always tell the truth) are numbered from 1 to 2021. Each of them states: "at least $i$ triples of participants can be formed, including me, that contain exactly two knights," where $i$ is the number assigned to the person speaking. How many possible values are there for the number of scoundrels present at the conference?
(A) 1
(B) 3
(C) 4
(D) 2021
(E) 2022
6. The answer is (B). We divide into cases based on the number of knights:

(a) The configuration with 0 knights, i.e., all rogues, is certainly valid.

(b) A configuration with exactly 1 knight is impossible, because this knight would claim to be able to find at least one other knight.

(c) If there are exactly 2 knights, the sequence $C C F F \ldots$ is valid: indeed, for the first two knights, it is possible to form a trio with any one of the 2019 rogues; moreover, each rogue has exactly one available trio, so he is indeed lying.

(d) With 3 knights, the argument is entirely analogous, and the sequence $C C C F F \ldots$ is valid.

(e) If there are at least 4 knights, observe first that there is at least one rogue, since there is at least one "CCF" trio. Then, each rogue can find at least $\left(\begin{array}{l}4 \\ 2\end{array}\right)=6$ trios with him and two knights, so the first six positions can only be occupied by knights. Repeating this reasoning, we find that the rogues must be after the position $\left(\begin{array}{l}6 \\ 2\end{array}\right)=$ 15, and then after the position $\left(\begin{array}{c}15 \\ 2\end{array}\right)=105$, and finally that the rogues can only occupy positions after $\left(\begin{array}{c}105 \\ 2\end{array}\right)>2021$; we would thus have a configuration of all knights, which, as stated at the beginning, is impossible.

In conclusion, the possible values for the number of rogues are only 2021, 2019, and 2018.