The ages of Mr. and Mrs. Fibonacci are both two-digit numbers. If Mr. Fibonacci’s age can be formed
by reversing the digits of Mrs. Fibonacci’s age, find the smallest possible positive difference between
their ages.

1. Let Mr. Fibonacci's age be represented by the two-digit number \(10a + b\), where \(a\) and \(b\) are the tens and units digits, respectively.
2. Let Mrs. Fibonacci's age be represented by the two-digit number \(10b + a\), where \(b\) and \(a\) are the tens and units digits, respectively.
3. We need to find the smallest possible positive difference between their ages, which is given by:
   \[
   |(10a + b) - (10b + a)|
   \]
4. Simplify the expression inside the absolute value:
   \[
   |(10a + b) - (10b + a)| = |10a + b - 10b - a| = |9a - 9b| = 9|a - b|
   \]
5. To minimize the positive difference, we need to minimize \(|a - b|\). Since \(a\) and \(b\) are digits (0 through 9), the smallest non-zero value for \(|a - b|\) is 1.
6. Therefore, the smallest possible positive difference is:
   \[
   9 \times 1 = 9
   \]
7. To verify, consider the ages where \(|a - b| = 1\). For example, if \(a = 4\) and \(b = 5\), then Mr. Fibonacci's age is 45 and Mrs. Fibonacci's age is 54. The difference is:
   \[
   |54 - 45| = 9
   \]

The final answer is \(\boxed{9}\)