Let $n\ge 3$ be a positive integer. Amy wrote all the integers from $1$ to $n^2$ on the $n\times n$ grid, so that each cell contains exactly one number. For $i=1,2,\cdots ,n^2-1$, the cell containing $i$ shares a common side with the cell containing $i+1$. Each turn, Bred can choose one cell, and check what number is written. Bred wants to know where $1$ is written by less than $3n$ turns. Determine whether $n$ such that Bred can always achieve his goal is infinite.
1. **Base Case Verification:**
   - For \( k = 1 \), \( n = 2^1 - 1 = 1 \). This is trivial since the grid is \( 1 \times 1 \) and Bred can find the number 1 in one turn, which is less than \( 3n = 3 \times 1 = 3 \).

2. **Inductive Hypothesis:**
   - Assume that for \( n = 2^k - 1 \), Bred can find the number 1 in less than \( 3n \) turns.

3. **Inductive Step:**
   - We need to show that for \( n = 2^{k+1} - 1 \), Bred can find the number 1 in less than \( 3n \) turns.
   - Let \( n = 2m + 1 \) where \( m = 2^k - 1 \). The grid is now \((2m+1) \times (2m+1)\).

4. **Labeling the Grid:**
   - Label the grid \((i, j)\) where \( 0 \leq i, j \leq 2m \).
   - Let \( f(i, j) \) be the number written in the cell at row \( i \) and column \( j \).

5. **Initial Checks:**
   - Bred first checks the middle row \( f(m, j) \) for \( 0 \leq j \leq 2m \).
   - Let \( k \) be the index such that \( f(m, k) = \min_{0 \leq j \leq 2m} f(m, j) \).
   - Check whether \( f(m-1, k) < f(m, k) \).

6. **Case Analysis:**
   - If \( f(m-1, k) < f(m, k) \), then the number 1 must lie between rows \( 0 \) to \( m-1 \).
   - If \( f(m-1, k) \geq f(m, k) \), then the number 1 must lie between rows \( m+1 \) to \( 2m \).

7. **Further Reduction:**
   - Assume \( f(m-1, k) < f(m, k) \). Now check \( f(0, m), f(1, m), \ldots, f(m-1, m) \).
   - If \( f(j, m) > f(m-1, k) \) for all \( 0 \leq j \leq m-1 \), then 1 is in the subgrid \((i, j)\) where \( 0 \leq i, j \leq m-1 \).
   - Otherwise, find the minimum \( f(j, m) \) and check \( f(j, m-1) \).

8. **Reduction to Smaller Grid:**
   - This process reduces the problem from a \((2m+1) \times (2m+1)\) grid to an \( m \times m \) grid in \( 2m+1 + 1 + m + 1 = 3m + 3 \) moves.
   - Since \( m = 2^k - 1 \), we have \( 3m + 3 = 3(2^k - 1) + 3 = 3 \times 2^k \).

9. **Conclusion:**
   - By induction, Bred can always find the number 1 in less than \( 3n \) turns for \( n = 2^k - 1 \).

The final answer is \( \boxed{ n = 2^k - 1 } \) for \( k \geq 1 \).