Problem 1307

To solve the problem, we need to consider the different cases where the number $n$ is a 3-digit number and $n'$ is a divisor of $n$, but $n' \neq n$.

### Case 1: Numbers with zero digits at the end
In this case, $n$ can be written as $n = 100a + 10b$ where $a$ and $b$ are non-zero digits (1 to 9 for $a$ and 0 to 9 for $b$). The number $n'$ is obtained by removing the zero digit, so $n' = 10a + b$.

Since $n'$ is a divisor of $n$, we have:
$$n = 100a + 10b$$
$$n' = 10a + b$$

We need to check if $n'$ divides $n$:
$$\frac{n}{n'} = \frac{100a + 10b}{10a + b}$$

For $n'$ to be a divisor of $n$, the above fraction must be an integer. Since $n' \neq n$, we need to ensure that $10a + b \neq 100a + 10b$.

Let's check the number of such 3-digit numbers:
- $a$ can be any digit from 1 to 9 (9 choices).
- $b$ can be any digit from 0 to 9 (10 choices).

Thus, there are:
$$9 \times 10 = 90$$
such numbers.

### Case 2: Numbers with one zero in the middle
In this case, $n$ can be written as $n = 100a + 10 \cdot 0 + b = 100a + b$ where $a$ and $b$ are non-zero digits (1 to 9 for $a$ and 1 to 9 for $b$). The number $n'$ is obtained by removing the zero digit, so $n' = 10a + b$.

We need to check if $n'$ divides $n$:
$$n = 100a + b$$
$$n' = 10a + b$$

We need to check if $n'$ divides $n$:
$$\frac{n}{n'} = \frac{100a + b}{10a + b}$$

For $n'$ to be a divisor of $n$, the above fraction must be an integer. Simplifying the fraction:
$$\frac{100a + b}{10a + b} = 10 + \frac{90a}{10a + b}$$

For this to be an integer, $\frac{90a}{10a + b}$ must be an integer. We can check the possible values of $a$ and $b$ that satisfy this condition.

By checking all values of $a$ from 1 to 9 and $b$ from 1 to 9, we find the following pairs:
$$(a, b) = (1, 5), (1, 8), (4, 5)$$

Thus, there are 3 such numbers.

### Conclusion
Adding the numbers from both cases, we get:
$$90 + 3 = 93$$

The final answer is $\boxed{93}$.