1. Let the two-digit integer be represented as $\underline{a}\, \underline{b}$, where $a$ and $b$ are the tens and units digits, respectively. This can be expressed as $10a + b$.

2. When this number is multiplied by 9, the resulting number is $9(10a + b)$.

3. According to the problem, the resulting three-digit number is of the form $\underline{a}\, \underline{c}\, \underline{b}$, which can be expressed as $100a + 10c + b$.

4. Equating the two expressions, we get:
$$9(10a + b) = 100a + 10c + b$$

5. Expanding and simplifying the equation:
$$90a + 9b = 100a + 10c + b$$
$$90a + 9b - 100a - b = 10c$$
$$-10a + 8b = 10c$$
$$10c = 8b - 10a$$
$$c = \frac{8b - 10a}{10}$$
$$c = \frac{4b - 5a}{5}$$

6. For $c$ to be a digit (i.e., an integer between 0 and 9), $4b - 5a$ must be divisible by 5. Let $4b - 5a = 5k$ for some integer $k$.

7. Solving for $b$:
$$4b = 5a + 5k$$
$$b = \frac{5a + 5k}{4}$$
Since $b$ must be a digit (0 to 9), $5a + 5k$ must be divisible by 4.

8. Let’s test possible values of $a$ (since $a$ is a digit from 1 to 9):
- For $a = 1$:
$$5(1) + 5k = 5 + 5k$$
$$5 + 5k \equiv 0 \pmod{4}$$
$$5 \equiv -3 \pmod{4}$$
$$-3 + 5k \equiv 0 \pmod{4}$$
$$5k \equiv 3 \pmod{4}$$
$$k \equiv 3 \pmod{4}$$
The smallest $k$ that satisfies this is $k = 3$:
$$b = \frac{5(1) + 5(3)}{4} = \frac{5 + 15}{4} = 5$$
So, $a = 1$, $b = 5$, and $c = \frac{4(5) - 5(1)}{5} = 3$.

- For $a = 2$:
$$5(2) + 5k = 10 + 5k$$
$$10 + 5k \equiv 0 \pmod{4}$$
$$10 \equiv 2 \pmod{4}$$
$$2 + 5k \equiv 0 \pmod{4}$$
$$5k \equiv -2 \pmod{4}$$
$$k \equiv 2 \pmod{4}$$
The smallest $k$ that satisfies this is $k = 2$:
$$b = \frac{5(2) + 5(2)}{4} = \frac{10 + 10}{4} = 5$$
So, $a = 2$, $b = 5$, and $c = \frac{4(5) - 5(2)}{5} = 2$.

- For $a = 3$:
$$5(3) + 5k = 15 + 5k$$
$$15 + 5k \equiv 0 \pmod{4}$$
$$15 \equiv -1 \pmod{4}$$
$$-1 + 5k \equiv 0 \pmod{4}$$
$$5k \equiv 1 \pmod{4}$$
$$k \equiv 1 \pmod{4}$$
The smallest $k$ that satisfies this is $k = 1$:
$$b = \frac{5(3) + 5(1)}{4} = \frac{15 + 5}{4} = 5$$
So, $a = 3$, $b = 5$, and $c = \frac{4(5) - 5(3)}{5} = 1$.

- For $a = 4$:
$$5(4) + 5k = 20 + 5k$$
$$20 + 5k \equiv 0 \pmod{4}$$
$$20 \equiv 0 \pmod{4}$$
$$5k \equiv 0 \pmod{4}$$
$$k \equiv 0 \pmod{4}$$
The smallest $k$ that satisfies this is $k = 0$:
$$b = \frac{5(4) + 5(0)}{4} = \frac{20 + 0}{4} = 5$$
So, $a = 4$, $b = 5$, and $c = \frac{4(5) - 5(4)}{5} = 0$.

9. The possible values of $\underline{a}\, \underline{b}$ are 15, 25, 35, and 45.

10. Summing these values:
$$15 + 25 + 35 + 45 = 120$$

The final answer is $\boxed{120}$.