Let $n \geq 3$ be a fixed integer. We need to find the largest number $m$ for which it is not possible to paint $m$ beads on a circular necklace using $n$ colors such that each color appears at least once among any $n+1$ consecutive beads.

### Analysis

1. Understanding the Problem:

Given $m$ beads and $n$ colors, the condition is that in any segment of $n+1$ consecutive beads, each of the $n$ colors must appear at least once.

We are seeking the largest $m$ such that this condition cannot be satisfied.

2. Pattern Exploration:

If we imagine arranging $n+1$ beads consecutively in a valid way, each subset of $n+1$ beads must exhaust all $n$ colors at least once.

Suppose you try to construct such a sequence that wraps around the circle of beads, large enough that fulfilling the requirement cannot be guaranteed.

3. Conceiving a Counterexample:

Assume $m = n^2 - n - 1$. We will denote the beads as positions $0, 1, 2, \ldots, n^2 - n - 2$.

To paint the beads, first consider a hypothetical repeating cycle of length $n$ (making use of all colors), repeating around until it fills $n^2 - n - 1$ positions exactly.

$$\text{Cycle}: (C_1, C_2, \ldots, C_n)$$

Now we note that, because the number of beads minus the number of colors $n+1$ (when considering one additional cycle bead space) does not sufficiently allow for coverage by distinct cycles, we are always missing coverage at $n^2 - n - 1 + n = n^2 - 1$ which is one more bead when closed in a necklace fashion.

4. Verification by Overshadowing Requirement:

Let’s test if it is actually possible with $m = n^2 - n$:

If $m = n^2-n$, consider that any setup you make inherently leaves a gap at some point due to the principle of covering $n+1$ beads (by symmetry and counting argument, this is best described as creating an incomplete residue class partition under modulus $n$, yielding a surplus one less than completing the cycle when $n+1$ is distributed).

### Conclusion

Through combinatorics and consequences of $m = n^2 - n - 1$ as number of beads, such a coloring following the constraints cannot exist, therefore the largest such value of $m$ for which the task is not possible is:

$$\boxed{n^2-n-1}$$

Thus, placing $n^2 - n - 1$ beads in circular fashion does not allow any $n+1$ to have all colors without duplicating elements mid-cycle, lacking exhaustiveness.