Suppose $n\geq 3$ is an integer. There are $n$ grids on a circle. We put a stone in each grid. Find all positive integer $n$, such that we can perform the following operation $n-2$ times, and then there exists a grid with $n-1$ stones in it: 

$\bullet$ Pick a grid $A$ with at least one stone in it. And pick a positive integer $k\leq n-1$. Take all stones in the $k$-th grid after $A$ in anticlockwise direction. And put then in the $k$-th grid after $A$ in clockwise direction. 
1. **Labeling the Grids:**
   Let's label the grids as \(1, 2, 3, \ldots, n\) in a clockwise direction. Each grid initially contains one stone.

2. **Understanding the Operation:**
   The operation allows us to pick a grid \(A\) with at least one stone and a positive integer \(k \leq n-1\). We then take all stones from the \(k\)-th grid after \(A\) in the anticlockwise direction and move them to the \(k\)-th grid after \(A\) in the clockwise direction.

3. **Analyzing the Movement:**
   When a stone is moved from grid \(i\) to grid \(j\), the difference \(i - j\) must be even. This is because the operation involves moving stones symmetrically around the circle.

4. **Case 1: \(n\) is Even:**
   - If \(n\) is even, say \(n = 2m\), then the difference \(i - j\) being even implies that stones can only move between grids with the same parity (odd to odd or even to even).
   - This restriction means that it is impossible to collect \(n-1\) stones into a single grid because we cannot move stones between grids of different parities.

5. **Case 2: \(n\) is Odd:**
   - If \(n\) is odd, say \(n = 2m + 1\), then the difference \(i - j\) can be odd or even, allowing stones to move between any grids.
   - We can perform the operation \(n-2\) times to collect \(n-1\) stones into one grid. For example, we can move stones in a specific sequence to ensure that one grid accumulates \(n-1\) stones.

6. **Conclusion:**
   - For even \(n\), it is impossible to collect \(n-1\) stones into one grid.
   - For odd \(n\), it is possible to collect \(n-1\) stones into one grid.

The final answer is all odd \( \boxed{ n } \).