Problem 1204

1. Understanding the problem: We need to find the maximum number of students $n$ in a class such that among any six students, there are two who are not friends, and for any pair of students that are not friends, there is a student among the remaining four who is friends with both of them.

2. Example construction: Consider 5 subgraphs $A, B, C, D, E$. Each subgraph has 5 vertices, and any two vertices in the same subgraph are not connected. Any two vertices in different subgraphs are connected. This construction ensures that each student has at least $n-5$ friends.

3. Claim: Everyone has at least $n-5$ friends.
- Proof of the claim: Let $A$ be a vertex. If $A$ is friends with all other vertices, the claim is trivially true. Suppose $B$ is a vertex that is not friends with $A$. There cannot be 4 vertices that are not friends with both $A$ and $B$ because, by the problem's condition, for any pair of students that are not friends, there is a student among the remaining four who is friends with both. Therefore, there are at least $n-5$ vertices that are friends with both $A$ and $B$. This proves the claim.

4. Degree calculation: For all $i = 1, 2, \ldots, n$, we have $\deg(v_i) \geq n-5$. Thus, the total number of edges $|E|$ in the graph is:
$$|E| = \frac{\sum \deg(v_i)}{2} \geq \frac{n(n-5)}{2}$$

5. Application of Turan's Theorem: We do not have a 6-clique, hence by Turan's Theorem, the number of edges $|E|$ is bounded by:
$$|E| \leq \frac{4n^2}{10} = \frac{2n^2}{5}$$

6. Combining inequalities:
$$\frac{n(n-5)}{2} \leq \frac{2n^2}{5}$$
Simplifying this inequality:
$$n(n-5) \leq \frac{4n^2}{5}$$
$$5n(n-5) \leq 4n^2$$
$$5n^2 - 25n \leq 4n^2$$
$$n^2 - 25n \leq 0$$
$$n(n-25) \leq 0$$

7. Solving the inequality: The inequality $n(n-25) \leq 0$ holds when $0 \leq n \leq 25$. Since $n$ must be a positive integer, the maximum value of $n$ is 25.

Conclusion:
$$\boxed{25}$$