The decimal digits of a natural number $A$ form an increasing sequence (from left to right).  Find the sum of the digits of $9A$.
1. Let \( A \) be a natural number whose decimal digits form an increasing sequence. We can represent \( A \) as \( A = \overline{a_n a_{n-1} a_{n-2} \dots a_0}_{10} \), where \( a_n, a_{n-1}, \dots, a_0 \) are the digits of \( A \) and \( a_n < a_{n-1} < \cdots < a_0 \).

2. To find the sum of the digits of \( 9A \), we first express \( 9A \) in terms of \( A \):
   \[
   9A = 10A - A
   \]

3. Consider the decimal representation of \( 10A \). If \( A = \overline{a_n a_{n-1} a_{n-2} \dots a_0}_{10} \), then:
   \[
   10A = \overline{a_n a_{n-1} a_{n-2} \dots a_0 0}_{10}
   \]
   This shifts all digits of \( A \) one place to the left and appends a zero at the end.

4. Now, subtract \( A \) from \( 10A \):
   \[
   9A = 10A - A = \overline{a_n a_{n-1} a_{n-2} \dots a_0 0}_{10} - \overline{a_n a_{n-1} a_{n-2} \dots a_0}_{10}
   \]

5. Perform the subtraction digit by digit:
   \[
   \overline{a_n a_{n-1} a_{n-2} \dots a_0 0}_{10} - \overline{a_n a_{n-1} a_{n-2} \dots a_0}_{10} = \overline{(a_n)(a_{n-1} - a_n)(a_{n-2} - a_{n-1}) \dots (a_0 - a_1 - 1)(10 - a_0)}_{10}
   \]

6. Since the digits of \( A \) form an increasing sequence, each subtraction \( a_{i-1} - a_i \) results in a non-negative digit, and the last term \( 10 - a_0 \) is also a non-negative digit.

7. The sum of the digits of \( 9A \) can be calculated as follows:
   \[
   \sum_{i=1}^{n} (a_{i-1} - a_i) + (10 - a_0)
   \]

8. Notice that the sum of the differences \( (a_{i-1} - a_i) \) telescopes to \( a_n - a_0 \):
   \[
   (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \cdots + (a_1 - a_0) = a_n - a_0
   \]

9. Therefore, the sum of the digits of \( 9A \) is:
   \[
   a_n - a_0 + (10 - a_0) = a_n - a_0 + 10 - a_0 = a_n + 10 - 2a_0
   \]

10. Since \( a_n \) and \( a_0 \) are digits of \( A \) and \( a_n < a_0 \), the sum of the digits of \( 9A \) simplifies to:
    \[
    9
    \]

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