Answer: $18$

For convenience we shall call 'a student whose class number is even' an even student. We make the following observations.

(1) Every question is solved by at least one even student (as the teacher could choose all even students).

(2) Similarly, every question is solved by at least one of students $5$, $10$, $15$, $20$.

(3) Students $2$, $4$, $6$, $\ldots$, $18$ (i.e. all even students except student $20$) cannot solve all questions (since any choice of the teacher must include either all even students or all students whose numbers are multiples of $5$). By (1), there must be a question which was solved by student $20$ but not by any other even student.

(4) For the same reason in (3), there must be a question which was solved by student $10$ but not by any other even student.

(5) We claim that there exist at least two questions that were solved by student $2$ but not any other even student. Using the same argument as in (3), we can show that there is at least one such question. Now suppose there is exactly one such question $P$. Now neither student $10$ nor $20$ solved $P$, so by (2) we know that either student $5$ or $15$ solved $P$. Without loss of generality suppose student $5$ did it. Consider students $4$, $5$, $6$, $8$, $10$, $12$, $14$, $16$, $18$, $20$ (i.e. all even students but with student $2$ replaced by student $5$). Since student $2$ and student $15$ are missing, these students together did not solve all questions. That means there is a question $Q$ which cannot be solved by these students (note that $P \neq Q$ since student $5$ solved $P$). As student $2$ cannot solve $Q$ either (since $P$ is the only question solved by student $2$ but not any other even student), that means no even student could solve $Q$, which contradicts (1). This proves the claim.

(6) For the same reason in (5), for each of students $4$, $6$, $8$, $12$, $14$, $16$, $18$, there must be at least two questions that were solved by that student but not any other even student.

From (3), (4), (5) and (6), there are at least $1 + 1 + 2 + 2 \times 7 = 18$ questions, and they are clearly distinct (by considering which even student solved each question). On the other hand, we can construct a paper with exactly $18$ questions as follows:

* Question $10$ — solved by student $10$ only
* Question $20$ — solved by student $20$ only
* Question $x$ for each $x \in \{2, 4, 6, 8, 12, 14, 16, 18\}$ — solved by students $5$ and $x$ only
* Question $x'$ for each $x' \in \{2, 4, 6, 8, 12, 14, 16, 18\}$ — solved by students $15$ and $x$ only

It can be easily checked that all even students together solved all questions, all students whose numbers are multiples of $5$ together solved all questions, and that if not all even students are present and not all students whose numbers are multiples of $5$ are present, then at least one question is not solved by the students picked. The answer is thus $18$.