Find the natural number $A$ such that there are $A$ integer solutions to $x+y\geq A$ where $0\leq x \leq 6$ and $0\leq y \leq 7$. 

[i]Proposed by David Tang[/i]
1. We need to find the natural number \( A \) such that there are \( A \) integer solutions to the inequality \( x + y \geq A \) where \( 0 \leq x \leq 6 \) and \( 0 \leq y \leq 7 \).

2. Let's consider the total number of integer pairs \((x, y)\) that satisfy the given constraints. The total number of pairs is:
   \[
   (6 - 0 + 1) \times (7 - 0 + 1) = 7 \times 8 = 56
   \]

3. We need to count the number of pairs \((x, y)\) that satisfy \( x + y \geq A \). To do this, we will count the number of pairs that do not satisfy the inequality \( x + y < A \) and subtract this from the total number of pairs.

4. For a given \( A \), the pairs \((x, y)\) that satisfy \( x + y < A \) form a triangular region in the coordinate plane. The number of such pairs is given by the sum of the first \( A-1 \) integers:
   \[
   \sum_{k=0}^{A-1} k = \frac{(A-1)A}{2}
   \]

5. We need to find \( A \) such that the number of pairs satisfying \( x + y \geq A \) is equal to \( A \). Therefore, we need:
   \[
   56 - \frac{(A-1)A}{2} = A
   \]

6. Solving the equation:
   \[
   56 - \frac{(A-1)A}{2} = A
   \]
   \[
   56 = A + \frac{(A-1)A}{2}
   \]
   \[
   56 = A + \frac{A^2 - A}{2}
   \]
   \[
   112 = 2A + A^2 - A
   \]
   \[
   A^2 + A - 112 = 0
   \]

7. Solving the quadratic equation \( A^2 + A - 112 = 0 \) using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   A = \frac{-1 \pm \sqrt{1 + 4 \cdot 112}}{2}
   \]
   \[
   A = \frac{-1 \pm \sqrt{1 + 448}}{2}
   \]
   \[
   A = \frac{-1 \pm \sqrt{449}}{2}
   \]

8. Since \( A \) must be a natural number, we take the positive root:
   \[
   A = \frac{-1 + \sqrt{449}}{2}
   \]

9. Approximating \(\sqrt{449}\):
   \[
   \sqrt{449} \approx 21.2
   \]
   \[
   A \approx \frac{-1 + 21.2}{2} \approx \frac{20.2}{2} \approx 10.1
   \]

10. Since \( A \) must be a natural number, we round to the nearest natural number:
   \[
   A = 10
   \]

11. Verifying, we find that the integer solutions to \( x + y \geq 10 \) where \( 0 \leq x \leq 6 \) and \( 0 \leq y \leq 7 \) are indeed 10 pairs: \((3,7), (4,6), (4,7), (5,5), (5,6), (5,7), (6,4), (6,5), (6,6), (6,7)\).

The final answer is \(\boxed{10}\).