Find the smallest positive integer, such that the sum and the product of it and $10$ are both square numbers. 
To find the smallest positive integer \( a \) such that both the sum \( 10 + a \) and the product \( 10a \) are perfect squares, we can proceed as follows:

1. **Set up the equations:**
   Let \( 10 + a = b^2 \) and \( 10a = c^2 \), where \( b \) and \( c \) are integers.

2. **Express \( a \) in terms of \( b \):**
   From \( 10 + a = b^2 \), we can solve for \( a \):
   \[
   a = b^2 - 10
   \]

3. **Substitute \( a \) into the second equation:**
   Substitute \( a = b^2 - 10 \) into \( 10a = c^2 \):
   \[
   10(b^2 - 10) = c^2
   \]
   Simplify this to:
   \[
   10b^2 - 100 = c^2
   \]
   \[
   c^2 = 10b^2 - 100
   \]

4. **Analyze the equation \( c^2 = 10b^2 - 100 \):**
   We need \( c^2 \) to be a perfect square. Notice that \( 10b^2 - 100 \) must be a perfect square. Let \( c = 10k \) for some integer \( k \), then:
   \[
   (10k)^2 = 10b^2 - 100
   \]
   \[
   100k^2 = 10b^2 - 100
   \]
   \[
   10b^2 = 100k^2 + 100
   \]
   \[
   b^2 = 10k^2 + 10
   \]

5. **Solve for \( b \) and \( k \):**
   We need \( b^2 = 10k^2 + 10 \) to be a perfect square. Let's test small values of \( k \):

   - For \( k = 1 \):
     \[
     b^2 = 10(1)^2 + 10 = 20 \quad \text{(not a perfect square)}
     \]

   - For \( k = 2 \):
     \[
     b^2 = 10(2)^2 + 10 = 50 \quad \text{(not a perfect square)}
     \]

   - For \( k = 3 \):
     \[
     b^2 = 10(3)^2 + 10 = 100 \quad \text{(a perfect square, since } b = 10)
     \]

6. **Find \( a \):**
   Since \( b = 10 \) and \( a = b^2 - 10 \):
   \[
   a = 10^2 - 10 = 100 - 10 = 90
   \]

Thus, the smallest positive integer \( a \) such that both \( 10 + a \) and \( 10a \) are perfect squares is \( \boxed{90} \).