To solve the problem, we need to find all positive integers $n$ such that the decimal representation of $n$, denoted as $\overline{a_m a_{m-1} \ldots a_1}$, satisfies the equation:
$$n = (a_m + 1)(a_{m-1} + 1) \cdots (a_1 + 1)$$

1. **Express $n$ in terms of its digits:**
$$n = \sum_{i=1}^{m} 10^{i-1} a_i$$
where $a_i$ are the digits of $n$.

2. Set up the given equation:
$$\sum_{i=1}^{m} 10^{i-1} a_i = \prod_{i=1}^{m} (a_i + 1)$$

3. Rearrange the equation:
$$\sum_{i=2}^{m} (10^{i-1} - \prod_{j=1}^{i-1} (a_j + 1)) a_i = 1$$
This equation implies that all summands are nonnegative, and one of them must be 1 while the others are 0.

4. **Consider the case $m = 2$:**
$$n = 10a_2 + a_1$$
$$n = (a_2 + 1)(a_1 + 1)$$
We need to find $a_1$ and $a_2$ such that:
$$10a_2 + a_1 = (a_2 + 1)(a_1 + 1)$$
Expanding and simplifying:
$$10a_2 + a_1 = a_2 a_1 + a_2 + a_1 + 1$$
$$10a_2 = a_2 a_1 + a_2 + 1$$
$$9a_2 = a_2 a_1 + 1$$
$$9a_2 - 1 = a_2 a_1$$
$$a_1 = \frac{9a_2 - 1}{a_2}$$
For $a_1$ to be an integer, $9a_2 - 1$ must be divisible by $a_2$. Testing values of $a_2$ from 1 to 9, we find:
$$a_2 = 2 \implies a_1 = \frac{9 \cdot 2 - 1}{2} = \frac{18 - 1}{2} = \frac{17}{2} \quad \text{(not an integer)}$$
$$a_2 = 1 \implies a_1 = \frac{9 \cdot 1 - 1}{1} = 8$$
Thus, $n = 10 \cdot 1 + 8 = 18$.

5. **Consider the case $m \geq 3$:**
If $m \geq 3$, then:
$$\prod_{j=1}^{m-1} (a_j + 1) = 10^{m-1}$$
This implies:
$$a_1 = a_2 = \cdots = a_{m-1} = 9$$
However, this is impossible because the product of digits $a_i + 1$ would be too large to match the sum of the digits multiplied by powers of 10.

Therefore, the only solution is when $m = 2$ and $n = 18$.

The final answer is $\boxed{18}$.