The smallest $n$ is 911. We divide the proof into two parts:

(1) We first prove that $n = 911$ satisfies the conditions. If each student answers at least 3 questions correctly, then for any student there are $\binom{15}{3} = 455$ ways for him to have exactly 3 correct answers. If there are 911 students participating in the contest, it follows from the pigeonhole principle that there are at least 3 students having 3 identical correct answers.

If there is a student $X$ whose score is not more than 2, then the number of remaining students with a score not more than 3 cannot exceed 10; otherwise, pick any 11 of these students together with $X$, and the sum of their total scores is less than 36 points. Then there are more than $911 - 11 = 900$ students in the rest, such that each of them has a score less than 4. Since $\binom{4}{3} = 4$, and $4 \times 900 > 455 \times 2$, there are at least 3 students answering 3 identical questions correctly.

(2) There are 910 students participating in the contest. Divide them into $455 = \binom{15}{3}$ groups, and there are exactly 2 students in each group. In each group, both students have the identical answers with only 3 correct answers indexed by the group label.