A square $ n \times n $, ($ n> 2 $) contains nonzero real numbers. It is known that every number is exactly $ k $ times smaller than the sum of all the numbers in its row or sum of all number in its column. For which real numbers $ k $ is this possible?
1. **Define the problem mathematically:**
   Let \( A \) be an \( n \times n \) matrix with nonzero real numbers. Let \( a_{ij} \) be the element in the \( i \)-th row and \( j \)-th column of \( A \). According to the problem, each element \( a_{ij} \) is \( k \) times smaller than the sum of the numbers in its row and column, excluding itself.

2. **Express the condition mathematically:**
   For each element \( a_{ij} \), the sum of the elements in the \( i \)-th row excluding \( a_{ij} \) is:
   \[
   S_{i} = \sum_{m=1, m \neq j}^{n} a_{im}
   \]
   Similarly, the sum of the elements in the \( j \)-th column excluding \( a_{ij} \) is:
   \[
   S_{j} = \sum_{m=1, m \neq i}^{n} a_{mj}
   \]
   The total sum of the elements in the cross (row and column excluding \( a_{ij} \)) is:
   \[
   S_{ij} = S_{i} + S_{j}
   \]
   According to the problem, we have:
   \[
   a_{ij} = \frac{1}{k} S_{ij}
   \]

3. **Sum all elements in the matrix:**
   Let \( T \) be the total sum of all elements in the matrix:
   \[
   T = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij}
   \]

4. **Express the sum of the cross sums:**
   The sum of all cross sums \( S_{ij} \) for all elements \( a_{ij} \) in the matrix can be written as:
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} S_{ij} = (n-1) \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} + (n-1) \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} - (n-1)^2 \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij}
   \]
   Simplifying, we get:
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} S_{ij} = 2(n-1)T - (n-1)^2 T
   \]

5. **Relate the sums to \( k \):**
   Using the condition \( a_{ij} = \frac{1}{k} S_{ij} \), we sum over all elements:
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} = \frac{1}{k} \sum_{i=1}^{n} \sum_{j=1}^{n} S_{ij}
   \]
   Substituting the expression for \( \sum_{i=1}^{n} \sum_{j=1}^{n} S_{ij} \):
   \[
   T = \frac{1}{k} \left( 2(n-1)T - (n-1)^2 T \right)
   \]
   Simplifying, we get:
   \[
   T = \frac{1}{k} \left( 2(n-1) - (n-1)^2 \right) T
   \]
   Since \( T \neq 0 \), we can divide both sides by \( T \):
   \[
   1 = \frac{1}{k} \left( 2(n-1) - (n-1)^2 \right)
   \]
   Solving for \( k \):
   \[
   k = \frac{1}{2(n-1) - (n-1)^2}
   \]

6. **Simplify the expression for \( k \):**
   \[
   k = \frac{1}{2n - 2 - n^2 + 2n - 1} = \frac{1}{-n^2 + 4n - 3}
   \]

Conclusion:
\[
k = \frac{1}{-n^2 + 4n - 3}
\]

The final answer is \( \boxed{ k = \frac{1}{-n^2 + 4n - 3} } \)