Example 1 In an exam, there are 5 multiple-choice questions, each with 4 different options, and each person selects exactly 1 option for each question. In 2000 answer sheets, it is found that there exists an $n$, such that in any $n$ answer sheets, there are 4 sheets where any 2 sheets have at most 3 answers the same. Find the minimum possible value of $n$.
(2000, China Mathematical Olympiad)
Solution: Let the 4 possible answers for each question be denoted as $1, 2, 3, 4$, and the answers on each test paper be denoted as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. Let $\{(1, h, i, j, k), (2, h, i, j, k), (3, h, i, j, k), (4, h, i, j, k)\}, h, i, j, k = 1, 2, 3, 4$, yielding a total of 256 quadruples.

Since $2000 = 256 \times 7 + 208$, by the pigeonhole principle, there are 8 test papers with answers belonging to the same quadruple. After removing these 8 test papers, there are still 8 test papers with answers belonging to the same quadruple among the remaining 1992 test papers; after removing these 8 test papers, there are still 8 test papers with answers belonging to the same quadruple among the remaining 1984 test papers. After removing these 8 test papers, a total of 24 test papers have been removed. Among these 24 test papers, any 4 of them have at least 2 with answers belonging to the same quadruple, which of course does not

meet the requirements of the problem. Therefore, $n \geqslant 25$.
Next, we prove that $n$ can be 25.
Let $S = \{(g, h, i, j, k) \mid g + h + i + j + k \equiv 0 \pmod{4}, g, h, i, j, k \in \{1,2,3,4\}\}$. Then $|S| = 256$, and any 2 answers in $S$ have at most 3 questions the same. Removing 6 elements from $S$, when each of the remaining 250 answers is chosen by exactly 8 people, a total of 2000 test papers are obtained, and among these 25 answers, there are always 4 that are different. Since they are all in $S$, they of course meet the requirements of the problem. This shows that $n = 25$ meets the requirements of the problem.
In summary, the minimum possible value of $n$ is 25.