Non-zero numbers are arranged in $n \times n$ square ($n>2$). Every number is exactly $k$ times less than the sum of all the other numbers in the same cross (i.e., $2n-2$ numbers written in the same row or column with this number). 
Find all possible $k$.

[i]Proposed by D. Rostovsky, A. Khrabrov, S. Berlov [/i]
1. **Understanding the Problem:**
   We are given an \( n \times n \) grid where each number is non-zero and every number is exactly \( k \) times less than the sum of all other numbers in the same cross (i.e., the \( 2n-2 \) numbers in the same row or column as this number). We need to find all possible values of \( k \).

2. **Analyzing the Condition:**
   Let \( a_{ij} \) be the number in the \( i \)-th row and \( j \)-th column. According to the problem, we have:
   \[
   a_{ij} = \frac{1}{k} \left( \sum_{m=1}^{n} a_{im} + \sum_{m=1}^{n} a_{mj} - 2a_{ij} \right)
   \]
   Simplifying, we get:
   \[
   a_{ij} = \frac{1}{k} \left( \sum_{m=1}^{n} a_{im} + \sum_{m=1}^{n} a_{mj} - 2a_{ij} \right)
   \]
   \[
   ka_{ij} = \sum_{m=1}^{n} a_{im} + \sum_{m=1}^{n} a_{mj} - 2a_{ij}
   \]
   \[
   (k+2)a_{ij} = \sum_{m=1}^{n} a_{im} + \sum_{m=1}^{n} a_{mj}
   \]

3. **Case Analysis:**
   - **Case \( k = 2n-2 \):**
     If all numbers in the table are equal, say \( a \), then:
     \[
     a = \frac{1}{2n-2} \left( (n-1)a + (n-1)a \right) = \frac{2(n-1)a}{2n-2} = a
     \]
     This is consistent, so \( k = 2n-2 \) is a solution.

   - **Case \( k = n-2 \):**
     Suppose all numbers in the first row are \( n-1 \) and all other numbers are \(-1\). For any \( a_{ij} \) in the first row:
     \[
     a_{ij} = n-1
     \]
     The sum of the other numbers in the same row and column is:
     \[
     (n-1) + (n-1) - 2(n-1) = 0
     \]
     This satisfies the condition, so \( k = n-2 \) is a solution.

   - **Case \( k = -2 \):**
     If the sum in each line (row or column) is zero, we can fill the inner \((n-1) \times (n-1)\) table randomly and determine the other numbers uniquely. This ensures all numbers are non-zero by choosing the original \((n-1)^2\) numbers to be positive. Thus, \( k = -2 \) is a solution.

4. **Contradiction for Other \( k \):**
   Assume \( k \ne 2n-2, n-2, -2 \). Let \( S \) be the sum of all numbers in the table. Using the hypothesis for all numbers and \( k \ne 2n-2 \), we get \( S = 0 \). Consider any \( a_1, \ldots, a_n \) forming a perfect matching \( b_1, \ldots, b_n \) in a straight line. Using \( k \ne -2, n-2 \) respectively yields:
   \[
   \sum a_i = \sum b_i = 0
   \]
   But \( \sum b_i = 0 \) forces \( k = -2 \), which is a contradiction.

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