Mongolian Mathematical Olympiad

Answer: 25.
Let's denote the number of students in the class as $N$. To begin, we show that it is possible for $N$ to equal 25. The class is divided into five groups, with each group consisting of five students who are not friends with each other. Furthermore, any two different groups of two children are friends. If we select a group of six students, there will always be at least two children from the same group who are not friends. As a result, among these children, there will be at least one child who is not part of their respective groups but is friends with them.

Next, we prove that $N$ is less than or equal to 25. Let's assume, by contradiction, that $N \ge 26$. We select a student, denoted as $a_1$, who has at least $N-5$ friends. If $a_1$ does not have at least five non-friends, it would contradict our assumption. We then choose another student, denoted as $a_2$, who is a friend of $a_1$. By following this pattern, we find that students $a_1, a_2, \dots, a_5$ have at least $N-25 \ge 1$ common friends. This directly contradicts the given condition that there are no pairs of non-friend students among any group of six students.