Let $n$ be an integer greater than $1$. A certain school has $1+2+\dots+n$ students and $n$ classrooms, with capacities for $1, 2, \dots, n$ people, respectively. The kids play a game in $k$ rounds as follows: in each round, when the bell rings, the students distribute themselves among the classrooms in such a way that they don't exceed the room capacities, and if two students shared a classroom in a previous round, they cannot do it anymore in the current round. For each $n$, determine the greatest possible value of $k$.

[i]Proposed by Victor Domínguez[/i]
1. **Understanding the Problem:**
   We need to determine the greatest possible value of \( k \) for which students can be distributed in \( n \) classrooms over \( k \) rounds, such that no two students share a classroom more than once. The classrooms have capacities \( 1, 2, \ldots, n \) respectively, and the total number of students is \( 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \).

2. **Initial Setup:**
   Let \( k = f(n) \). We need to show that \( f(n) \leq 3 \).

3. **First Round Distribution:**
   Suppose in the first round, students \( a_1, a_2, \ldots, a_n \) are in classroom \( n \). Let the set of these students be \( A \), and let \( B \) be the set of the other students. The students in \( B \) will be distributed in the first \( n-1 \) classrooms.

4. **Subsequent Rounds:**
   In the next rounds, all students in \( A \) must be in different classrooms. This means that for each classroom, they "steal" one spot in it. Therefore, the students in \( B \) distribute in rooms with capacities \( 1, 2, \ldots, n-1 \).

5. **Recursive Argument:**
   Since the students in \( B \) now face a similar problem but with \( n-1 \) classrooms, we can recursively apply the same argument. This gives us \( f(n) \leq f(n-1) \).

6. **Base Case:**
   For \( n = 2 \), we have two classrooms with capacities 1 and 2, and a total of 3 students. It is easy to see that \( f(2) = 3 \) because:
   - In the first round, one student is in the first classroom, and two students are in the second classroom.
   - In the second round, the student from the first classroom moves to the second classroom, and the two students from the second classroom move to the first and second classrooms respectively.
   - In the third round, the students can be rearranged again without any two students sharing a classroom more than once.

7. **Conclusion:**
   By the recursive argument and the base case, we have \( f(n) \leq f(n-1) \leq \cdots \leq f(2) = 3 \). Therefore, \( f(n) \leq 3 \).

\(\blacksquare\)

The final answer is \( \boxed{ f(n) \leq 3 } \).