There are $\displaystyle{2n}$ students in a school $\displaystyle{\left( {n \in {\Bbb N},n \geqslant 2} \right)}$. Each week $\displaystyle{n}$ students go on a trip. After several trips the following condition was fulfiled: every two students were together on at least one trip. What is the minimum number of trips needed for this to happen?

[i]Proposed by Oleksandr Rybak, Kiev, Ukraine.[/i]
To solve the problem, we need to determine the minimum number of trips required such that every pair of students is together on at least one trip. We are given that there are \(2n\) students and each trip consists of \(n\) students.

1. **Establishing the minimum number of trips:**
   - Each student must be paired with every other student at least once.
   - There are \(\binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1)\) pairs of students.
   - Each trip can accommodate \(\binom{n}{2} = \frac{n(n-1)}{2}\) pairs of students.

2. **Calculating the lower bound:**
   - Let \(k\) be the number of trips needed.
   - Each student appears in \(k\) trips, and each trip contains \(n\) students.
   - Each student pairs with \(n-1\) other students in each trip.
   - Therefore, each student pairs with \((n-1)k\) other students across all trips.
   - Since each student must pair with \(2n-1\) other students, we have:
     \[
     (n-1)k \geq 2n-1
     \]
   - Solving for \(k\):
     \[
     k \geq \frac{2n-1}{n-1}
     \]
   - For \(n \geq 2\), this simplifies to:
     \[
     k \geq 2 + \frac{1}{n-1}
     \]
   - Since \(k\) must be an integer, the minimum \(k\) is 3 when \(n=2\), and for larger \(n\), \(k\) must be at least 6.

3. **Constructing the solution for specific \(n\):**
   - For \(n=2\), the set of students is \(\{x, y, z, t\}\). The trips can be:
     \[
     \{x, y\}, \{x, z\}, \{x, t\}, \{y, z\}, \{y, t\}, \{z, t\}
     \]
   - For \(n=3\), the set of students is \(\{a, b, c, d, e, f\}\). The trips can be:
     \[
     \{a, b, c\}, \{a, b, d\}, \{c, d, e\}, \{c, d, f\}, \{e, f, a\}, \{e, f, b\}
     \]

4. **Generalizing the solution:**
   - For larger \(n\), we can use the above models and merge them appropriately.
   - For example, for \(n=5\), we can combine the models for \(n=2\) and \(n=3\).

5. **Conclusion:**
   - The minimum number of trips needed is 6 for any \(n \geq 2\).

The final answer is \(\boxed{6}\).