24.3.11  For a competition with 4 multiple-choice questions, each with $A, B, C$ as options, for any three participants, there is at least one question where their answers are all different. How many people can participate in the competition at most?
Using the recursive method, let $f(n)$ represent the maximum number of participants when there are $n$ questions, then $f(1)=3$.
If a question makes the answers of three people different, we say this question can distinguish these three people. By the pigeonhole principle, for the first question, there must be an option chosen by at most $\left[\frac{f(n)}{3}\right]$ people. Therefore, any three of the remaining $f(n)-\left[\frac{f(n)}{3}\right]$ people cannot be distinguished by the first question. So, the remaining $n-1$ questions can distinguish these people, hence $f(n-1) \geqslant f(n)-\left[\frac{f(n)}{3}\right] \geqslant \frac{2}{3} f(n)$, which means $f(n) \leqslant \frac{3}{2} f(n-1)$. Therefore, $f(2) \leqslant 4, f(3) \leqslant 6, f(4) \leqslant 9$. An example with 9 people is given as follows:
\begin{tabular}{lllll} 
& I & II & III & IV \\
1 & A & A & A & A \\
2 & A & B & B & B \\
3 & A & C & C & C \\
4 & B & A & C & B \\
5 & B & B & A & C \\
6 & B & C & B & A \\
7 & C & A & B & C \\
8 & C & B & C & A \\
9 & C & C & A & B
\end{tabular}

Therefore, at least 9 people can participate in the competition.