6. (8 points) At Innopolis University, $2 n(n>1)$ students are enrolled. It turned out that any $2 n-2$ of them can be divided into $n-1$ pairs of acquaintances. What is the smallest number of pairs of acquaintances that can exist among all students.
Solution: Let there be a student (let's call him Bob) who has no more than two acquaintances. Then consider a group consisting of Bob and another $2 n-3$ students who are not acquainted with Bob. We have found a group that contradicts the problem's condition (no one can form a pair with Bob). Therefore, we have proven that each student has at least 3 acquaintances. Then the number of pairs of acquaintances is at least $\frac{3 \cdot 2 n}{2}=3 n$.

Let's provide an example where there are exactly $3 n$ acquaintances. Seat all students around a round table and say that each student is acquainted with their immediate neighbors and the student sitting directly opposite. Consider any group of $2 n-2$ students. If we remove two students standing opposite each other, the remaining can be divided into $n-1$ pairs standing opposite. If we remove two students not standing opposite each other, and if each of the two arcs formed by them has an even number of students, we will divide them into pairs of neighbors. If each arc has an odd number of students, on the smaller arc we will divide all but one of the extreme students into pairs, and give the extreme student a pair with the opposite one. After this, as it is not difficult to check, the larger arc will split into two with an even number of students, and we can divide each into pairs of neighbors.

## Answers:

| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 153 | 303 | 228 | 183 | 213 | 243 | 273 | 198 | 258 | 288 | 318 |