Find all positive integers $ n $ with the following property: It is possible to fill a $ n \times n $ chessboard with one of arrows $ \uparrow, \downarrow, \leftarrow, \rightarrow $ such that 

1. Start from any grid, if we follows the arrows, then we will eventually go back to the start point.

2. For every row, except the first and the last, the number of $ \uparrow $ and the number of $ \downarrow $ are the same. 

3. For every column, except the first and the last, the number of $ \leftarrow $ and the number of $ \rightarrow $ are the same.
1. **Claim**: The only positive integer \( n \) that satisfies the given conditions is \( n = 2 \).

2. **Verification for \( n = 2 \)**:
   - Consider a \( 2 \times 2 \) chessboard. We can fill it with arrows such that each cell points to another cell, forming a cycle. For example:
     \[
     \begin{array}{|c|c|}
     \hline
     \rightarrow & \downarrow \\
     \hline
     \uparrow & \leftarrow \\
     \hline
     \end{array}
     \]
   - Starting from any cell and following the arrows, we will eventually return to the starting cell, satisfying condition 1.
   - Each row and each column has one \(\uparrow\) and one \(\downarrow\) (or \(\leftarrow\) and \(\rightarrow\)), satisfying conditions 2 and 3.

3. **General Case**:
   - Suppose \( n \) is an even integer, \( n = 2m \).
   - Label the columns \( 1, 2, \ldots, n \) from left to right.
   - Draw a directed edge from \( u \) to \( v \) if the arrow in square \( u \) points from \( u \) to \( v \). The directed edges form a union of disjoint cycles, implying \( 2 \mid n \).

4. **Cycle Analysis**:
   - Consider the cycles that enter/exit a column. The number of rightward-pointing arrows in column \( c \) must equal the number of leftward-pointing arrows in column \( c+1 \).
   - Condition 3 implies that the number of rightward arrows in each column (except the rightmost one) is the same, denoted as \( k \). Similarly, the number of leftward arrows in each column (except the leftmost one) is also \( k \).

5. **Arrow Counting**:
   - The total number of leftward arrows is \( (2m-1)k \).
   - Let \( l \) be the number of upward arrows in each row (except the topmost one). The total number of upward arrows is \( (2m-1)l \).
   - Therefore, the total number of arrows is \( (2m-1)(k + l) \cdot 2 \).

6. **Equality Condition**:
   - The total number of arrows must also equal \( 4m^2 \) (since each cell has one arrow and there are \( n^2 = (2m)^2 \) cells).
   - Thus, we have:
     \[
     (2m-1)(k + l) \cdot 2 = 4m^2
     \]
     Simplifying, we get:
     \[
     (2m-1)(k + l) = 2m^2
     \]

7. **Divisibility Condition**:
   - Since \( 2m-1 \) and \( 2m \) are coprime, \( 2m-1 \) must divide \( 2m^2 \).
   - The only integer solution to \( 2m-1 = 1 \) is \( m = 1 \), giving \( n = 2m = 2 \).

8. **Conclusion**:
   - Therefore, the only positive integer \( n \) that satisfies all the given conditions is \( n = 2 \).

The final answer is \( \boxed{ n = 2 } \).