A positive two-digit number is such that the product of the digits is 20. When the digits are reversed, the number so formed is greater than the original number by 9. Find the number.

Let the tens digit be $t$ and the units digit be $u$.
The original number can be expressed as $10t + u$.
The reversed number can be expressed as $10u + t$.

Condition 1: The product of the digits is 20.

Condition 2: When the digits are reversed, the new number is greater than the original number by 9.

Subtract $u$ and $t$ from both sides:

Divide by 9:

Substitute $u = t + 1$ into $t \times u = 20$:

Rearrange into a quadratic equation:

Factor the quadratic equation:

This gives two possible values for $t$:

Since $t$ is a digit of a positive two-digit number, it must be a positive integer between 1 and 9. Therefore, $t = 4$.

Using Equation (3), $u = t + 1$:

The tens digit is $t=4$ and the units digit is $u=5$.
The original number is $10t + u = 10(4) + 5 = 40 + 5 = 45$.

Check the conditions:
Product of digits: $4 \times 5 = 20$ (Correct)
Reversed number: 54.
Original number: 45.
Is $54 = 45 + 9$? Yes, $54 = 54$ (Correct)

The number is $\boxed{45}$.