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 $\boxed{9}$.