The maximum possible value of $n$ is $18$.

We first show that $n \ge 19$ is impossible. Let $A$, $B$, $C$ be the answers in each problem. For $n \ge 19$, by the pigeonhole principle, we may assume $\left\lfloor \frac{19}{3} \right\rfloor = 6$ students answer $A$ in problem 1. Then by the pigeonhole principle, we may assume $\left\lfloor \frac{6}{3} \right\rfloor = 2$ students answer $A$ in problem 2. These 2 students must have different answers to the remaining problems. Thus, we may assume their answers are

AAAAAA, ABBBBB, AACCCC.

Consider another student choosing answer $A$ in problem 1. By the pigeonhole principle, this student chooses the same choice in $\left\lfloor \frac{4}{3} \right\rfloor = 1$ of the problems among the remaining problems. This leads to a contradiction, as it is not possible to construct such a set of students.

Therefore, we must have $n \le 18$.

We now provide an example for $n = 18$. The following lists out the answers given by 18 students:

AAAAAA AABBCC ABCACB ACBCAB ABACBC ACCBBA
BBBBBB BBCCAA BCABAC BACABC BCBACA BAACCB
CCCCCC CCAABB CABCBA CBABCA CACBAB CBBAAC

It is routine to check that any two students in the same column share 0 same answer, while any two students not in the same column share 2 same answers. (Note that there is a cyclic symmetry in each column.)