In a pond there are $n \geq 3$ stones arranged in a circle. A princess wants to label the stones with the numbers $1, 2, \dots, n$ in some order and then place some toads on the stones. Once all the toads are located, they start jumping clockwise, according to the following rule: when a toad reaches the stone labeled with the number $k$, it waits for $k$ minutes and then jumps to the adjacent stone.

What is the greatest number of toads for which the princess can label the stones and place the toads in such a way that at no time are two toads occupying a stone at the same time?

[b]Note:[/b] A stone is considered occupied by two toads at the same time only if there are two toads that are on the stone for at least one minute.
1. **Claim**: The maximum number of toads that the princess can place is \(\left\lceil \frac{n}{2} \right\rceil\).

2. **Labeling the Stones**: Let \(a_1, a_2, \dots, a_n\) be the labels of the stones in order (after \(a_n\) comes \(a_1\) again). We consider \(a_{n+i} = a_i\) for the rest of the proof.

3. **Condition for No Collision**: If there is a toad on \(a_l\) and a toad on \(a_k\), they will never meet if and only if \(a_l + a_{l+1} + \dots + a_{k-1} \ge a_m\) for \(1 \le m \le n\).

4. **Proof of Condition**: 
   - If \(a_l + a_{l+1} + \dots + a_{k-1} < a_k\), the two toads will meet on \(a_k\) at the minute \(a_l + a_{l+1} + \dots + a_{k-1}\). Therefore, \(a_l + a_{l+1} + \dots + a_{k-1} \ge a_k\).
   - For any \(q \ge k\), if \(a_l + a_{l+1} + \dots + a_{k-1} + a_k + a_{k+1} + \dots + a_{q-1} < a_k + a_{k+1} + \dots + a_{q}\), the two toads will meet on \(a_q\) at the minute \(a_l + a_{l+1} + \dots + a_{q-1}\). Therefore, \(a_l + a_{l+1} + \dots + a_{q-1} \ge a_k + a_{k+1} + \dots + a_{q}\), which implies \(a_l + a_{l+1} + \dots + a_{k-1} \ge a_q\).

5. **Labeling Strategy**: The princess labels the stones in groups of two that add up to \(n\) in the following way:
   \[
   n, 1, n-1, 2, n-2, \dots, k, n-k, \dots
   \]
   If the princess places a toad on the first of each group, meaning on \(n\), and on each \(k\) with \(k < \frac{n}{2}\), this configuration works and has \(\left\lceil \frac{n}{2} \right\rceil\) toads. If \(n\) is even, she places \(\frac{n}{2}\) before \(n\) and the configuration still works. This is because the distance between any two toads is at least 2 stones (except for the case of the toad on \(n\)), and thanks to the configuration, the sum of the labels of these two stones is \(n\). Using the lemma, we conclude that no two toads are going to meet.

6. **Pigeonhole Principle**: To finish the problem, consider that the princess places \(\left\lceil \frac{n}{2} \right\rceil + 1\) toads. By the pigeonhole principle, this means there are at least two toads with a toad directly in front of them. Meaning there is a \(k\) such that there is a toad on \(a_k\) and a toad on \(a_{k+1}\) and \(a_k < n\). Using the lemma, we can conclude that these two toads will meet on the stone with the label \(n\), and hence it is impossible for the princess to place this many toads.

\(\blacksquare\)

The final answer is \( \boxed{ \left\lceil \frac{n}{2} \right\rceil } \).