Solution:
We do casework on the number of digits of $n$.

One digit. There are $4$ one-digit stacked integers: $1, 2, 3, 4$.

Two digits. Suppose $n = \overline{ab}$ is a two-digit integer. If $a < 5$ and $b < 5$, then the digits of $2 n$ are double the respective digits of $n$, so $n$ is stacked; there are $4 \cdot 5 = 20$ such $n$. Otherwise, since $2 n < 100$, we still must have $a < 5$, so $b \geq 5$. Then the last digit of $2 n$ is $2b - 10$, so $b \mid 2b - 10$, which implies that $b = 5$. Then the first digit of $2 n$ is $2a + 1$, which $a$ must divide, so $a = 1$. Thus, the only stacked $n$ with $b \geq 5$ is $15$. Adding that to the $20$ stacked numbers with $b < 5$ gives us $21$ two-digit stacked integers.

Three digits. Suppose $n = \overline{abc}$ is a three-digit integer. If $a, b$, and $c$ are all less than $5$, then the digits of $2 n$ are double the respective digits of $n$, so $n$ is stacked; there are $4 \cdot 5 \cdot 5 = 100$ such $n$. Otherwise, since $2 n < 1000$, we must have $a < 5$. We now casework on which of $b$ and $c$ are at least $5$.

- If $b \geq 5$ and $c \geq 5$, then the digits of $2 n$ are $2a + 1, 2b - 9$, and $2c - 10$ in order. Thus, $a \mid 2a + 1$, $b \mid 2b - 9$, and $c \mid 2c - 10$, which implies $a = 1$, $b = 9$, and $c = 5$. Thus $195$ is the only stacked number in this case.

- If $c \geq 5$ only, then $2 n = 200a + 2\overline{bc}$ has first digit $2a$ and last two digits $2\overline{bc}$, so $n$ is stacked if and only if $\overline{bc}$ to be stacked. Since $c \geq 5$, as proved before, the only such stacked $\overline{bc}$ is $15$, so we get $4$ stacked numbers in this case: $115, 215, 315$, and $415$.

- If $b \geq 5$ only, then $2 n$ has last digit $2c$ and first two digits $2\overline{ab}$, so $n$ is stacked if and only if $\overline{ab}$ to be stacked. As $b \geq 5$, similar to the previous case, the only such stacked $\overline{ab}$ is $\overline{ab} = 15$, so we get $5$ stacked numbers in this case: $150, 151, 152, 153$, and $154$.

Summing over all cases, there are $100 + 1 + 4 + 5 = 110$ three-digit stacked integers.

Our final answer is $4 + 21 + 110 = 135$.