We have filled a 10 × 10 grid with numbers +1 and −1. Find the maximum amount of k such that there are exactly k rows with positive sum and exactly k columns with negative sum.

To solve the problem, we need to analyze the structure of a \(10 \times 10\) grid filled with numbers \(+1\) and \(-1\).

1. **Definitions**:
   - Let \(k\) be the number of rows with a positive sum.
   - Consequently, there will be \(10 - k\) rows with a negative sum.
   - Similarly, \(k\) will also be the number of columns with a negative sum, leading to \(10 - k\) columns with a positive sum.

2. **Total Sum Calculation**:
   - The total sum of all the entries in the grid can be expressed as:
     \[
     S = (+1) \cdot a + (-1) \cdot b
     \]
     where \(a\) is the count of \(+1\)s and \(b\) is the count of \(-1\)s. Since the grid contains 100 entries, we have:
     \[
     a + b = 100
     \]

3. **Row and Column Sums**:
   - For each row with a positive sum, the number of \(+1\)s must exceed the number of \(-1\)s. Therefore, if a row has \(n_i\) \(+1\)s, then:
     \[
     n_i > 5 \quad \text{(for positive sum)}
     \]
   - Conversely, for rows with negative sums, the opposite must hold, so:
     \[
     n_i < 5 \quad \text{(for negative sum)}
     \]

4. **Balancing Rows and Columns**:
   - The total number of \(+1\)s in \(k\) rows and \(10 - k\) rows can be denoted as:
     \[
     \text{Total from } k \text{ rows} > 5k
     \]
     \[
     \text{Total from } (10 - k) \text{ rows} < 5(10 - k)
     \]
   - Thus, we can set up the inequality for the total number of \(+1\)s:
     \[
     a > 5k \quad \text{and} \quad a < 5(10 - k)
     \]
     Combining these:
     \[
     5k < a < 50 - 5k
     \]

5. **Combining Inequalities**:
   - Therefore:
     \[
     5k < 50 - 5k \implies 10k < 50 \implies k < 5
     \]
   - Since \(k\) must also be a non-negative integer, we find:
     \[
     k \leq 4
     \]

6. **Finding Maximum \(k\)**:
   - Testing \(k = 4\):
     - If \(k = 4\), we have \(4\) rows with positive sums and \(6\) rows with negative sums.
     - The count of \(+1\)s will need to satisfy:
       \[
       a > 5 \cdot 4 = 20 \quad \text{and} \quad a < 50 - 5 \cdot 4 = 30
       \]
       Thus, \(a\) can take values \(21, 22, \ldots, 29\).

   - The configuration is possible, ensuring that in the remaining columns (with \(10 - k = 6\)), there are enough \(-1\)s to maintain the conditions. 

In conclusion, the maximum \(k\) such that there are exactly \(k\) rows with a positive sum and exactly \(k\) columns with a negative sum is:

\[
\boxed{4}
\]

### Further Exploration
Would you like to explore:
1. Examples of specific configurations of the grid?
2. A different approach to the problem?
3. How varying \(k\) affects the sums?
4. Similar problems with different grid sizes?
5. Detailed proofs of the inequalities used? 

### Tip
When tackling problems involving combinatorial configurations, always consider the implications of constraints on sums and distributions within the grid.

Explore how to determine the maximum value of k in a 10x10 grid filled with +1 and -1 numbers. This solution involves combinatorial reasoning and inequalities, resulting in k being maximized at 4. Perfect for advanced high school mathematics students.

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