The answer is $\boxed{25}.$ First let us show that $25$ actually works. Note that there are $4^5 = 1024$ different possible combinations of answers that the students could have obtained. Now, simply take $2000$ students such that there are at most two students who obtained any combination of answers. We claim that this works for $n = 25.$ As any answer combination appears at most twice, these students must have at least $13$ distinct answer combinations. Call the answers for each question $0, 1, 2, 3,$ and for each answer combination consider the sum of the answers modulo $4.$ By Pigeonhole at least $4$ of these combinations will have this sum equal to one another, and it can be checked that four students with these combinations work. Arbitrarily select $3$ of the $4^4 = 256$ possible answer combinations of the first four question. The expected number of students whose first four answers coincide with one of these $3$ answer combinations is $\frac{3}{256} \cdot 200 > 23$, and so there is some triple of combinations such that the number of students whose first four answers coincide with one of them is at least $24.$ It can then be checked that these $24$ students violate the desired condition. So the answer is $\boxed{25}.$ $\square$