Determine the maximum possible number of points you can place in a rectangle with lengths $14$ and $28$ such that any two of those points are more than $10$ apart from each other.
1. **Partitioning the Rectangle:**
   - The rectangle has dimensions \(14 \times 28\).
   - We partition the rectangle into \(8\) squares, each of dimensions \(7 \times 7\).
   - If we place more than \(8\) points in the rectangle, by the pigeonhole principle, at least two points will lie in the same \(7 \times 7\) square.
   - The maximum distance between any two points in a \(7 \times 7\) square is the diagonal, which is \(\sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.899 < 10\).
   - Therefore, no two points in the same \(7 \times 7\) square can be more than \(10\) units apart.

2. **Placing the Points:**
   - Consider the rectangle with vertices \((0,0)\), \((28,0)\), \((28,14)\), and \((0,14)\).
   - Define \(x = 9 + \frac{1}{3}\) and \(y = 4 - \frac{1}{4}\).
   - Place \(8\) points at the following coordinates:
     \[
     (0,0), (x,y), (2x,0), (3x,y), (0,14-y), (x,14), (2x,14-y), (3x,14)
     \]
   - We need to verify that the distance between any two points is greater than \(10\).

3. **Distance Calculation:**
   - Calculate the distance between \((0,0)\) and \((x,y)\):
     \[
     \sqrt{x^2 + y^2} = \sqrt{\left(9 + \frac{1}{3}\right)^2 + \left(4 - \frac{1}{4}\right)^2} = \sqrt{\left(\frac{28}{3}\right)^2 + \left(\frac{15}{4}\right)^2} = \sqrt{\frac{784}{9} + \frac{225}{16}}
     \]
     \[
     = \sqrt{\frac{3136}{36} + \frac{50625}{576}} = \sqrt{\frac{3136 \cdot 16 + 50625 \cdot 9}{576}} = \sqrt{\frac{50176 + 455625}{576}} = \sqrt{\frac{505801}{576}} \approx 10.01
     \]
   - Since \(\sqrt{x^2 + y^2} > 10\), the distance between \((0,0)\) and \((x,y)\) is greater than \(10\).

4. **Verification of Other Distances:**
   - The distance between \((0,0)\) and \((2x,0)\) is:
     \[
     2x = 2 \left(9 + \frac{1}{3}\right) = 2 \cdot \frac{28}{3} = \frac{56}{3} \approx 18.67 > 10
     \]
   - The distance between \((0,0)\) and \((3x,y)\) is:
     \[
     \sqrt{(3x)^2 + y^2} = \sqrt{\left(3 \cdot \frac{28}{3}\right)^2 + \left(4 - \frac{1}{4}\right)^2} = \sqrt{28^2 + \left(\frac{15}{4}\right)^2} = \sqrt{784 + \frac{225}{16}} = \sqrt{784 + 14.0625} \approx 28.26 > 10
     \]
   - Similar calculations can be done for other pairs of points to verify that all distances are greater than \(10\).

Conclusion:
- We have shown that it is possible to place \(8\) points in the rectangle such that the distance between any two points is greater than \(10\).
- Therefore, the maximum number of points that can be placed in the rectangle under the given conditions is \(8\).

The final answer is \(\boxed{8}\)