Let $n\ge 3$ be an integer. Annalena has infinitely many cowbells in each of $n$ different colours. Given an integer $m \ge n + 1$ and a group of $m$ cows standing in a circle, she is tasked with tying one cowbell around the neck of every cow so that every group of $n + 1$ consecutive cows have cowbells of all the possible $n$ colours. Prove that there are only finitely many values of $m$ for which this is not possible and determine the largest such $m$ in terms of $n$.
1. **Restate the problem in terms of beads and colors:**
   We need to paint \( m \) beads on a circular necklace using \( n \) colors such that every group of \( n+1 \) consecutive beads contains all \( n \) colors. We aim to find the largest \( m \) for which this is not possible.

2. **Understand the constraints:**
   - We have \( n \) colors.
   - We need every group of \( n+1 \) consecutive beads to contain all \( n \) colors.
   - We need to find the largest \( m \) for which this is not possible.

3. **Consider the case when \( m = n+1 \):**
   - If \( m = n+1 \), we need to color \( n+1 \) beads such that every group of \( n+1 \) consecutive beads contains all \( n \) colors.
   - This is impossible because in a circular arrangement of \( n+1 \) beads, there will always be a repetition of at least one color, making it impossible to have all \( n \) colors in every group of \( n+1 \) consecutive beads.

4. **Generalize to larger \( m \):**
   - For \( m > n+1 \), we need to ensure that every group of \( n+1 \) consecutive beads contains all \( n \) colors.
   - If \( m = n+2 \), we can still find a configuration where every group of \( n+1 \) consecutive beads contains all \( n \) colors. For example, we can repeat a sequence of \( n \) colors followed by one additional color.
   - As \( m \) increases, it becomes easier to satisfy the condition because we have more beads to distribute the colors.

5. **Determine the largest \( m \) for which the task is not possible:**
   - The task is not possible for \( m = n+1 \) because we cannot distribute \( n \) colors among \( n+1 \) beads in a circular arrangement without repeating a color in every group of \( n+1 \) consecutive beads.
   - For \( m > n+1 \), it is possible to find a configuration that satisfies the condition.

6. **Conclusion:**
   - The largest value of \( m \) for which it is not possible to paint the beads as required is \( m = n+1 \).

The final answer is \( \boxed{ n+1 } \).