Problem 1219

1. Given that $n$ has exactly 12 positive divisors, we can write the number of divisors function as:
$$d(n) = 12$$
The number of divisors of $n$ can be expressed in terms of its prime factorization. If $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, then:
$$d(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$$
Since $d(n) = 12$, the possible factorizations of 12 are:
$$12 = 12 \times 1, \quad 6 \times 2, \quad 4 \times 3, \quad 3 \times 2 \times 2$$
This means $n$ could be of the form $p_1^{11}$, $p_1^5 p_2$, $p_1^3 p_2^2$, or $p_1^2 p_2 p_3$.

2. Let $d_4$ be the 4th smallest divisor of $n$. We are given $m = d_4 - 1$ and $d_m = (d_1 + d_2 + d_4) d_8$.

3. Since $d_1 = 1$, we have:
$$d_m = (1 + d_2 + d_4) d_8$$

4. We need to determine $d_4$. Notice that $d_4$ is the 4th smallest divisor, so it is relatively small. Let's consider the possible forms of $n$ and their divisors.

5. If $n = p_1^2 p_2 p_3$, then the divisors are:
$$1, p_1, p_2, p_3, p_1^2, p_1 p_2, p_1 p_3, p_2 p_3, p_1^2 p_2, p_1^2 p_3, p_1 p_2 p_3, p_1^2 p_2 p_3$$
We need to find $p_1, p_2, p_3$ such that the 4th smallest divisor $d_4$ fits the given conditions.

6. Let's assume $d_4 = 13$. Then:
$$m = d_4 - 1 = 13 - 1 = 12$$
So, $d_{12} = n$.

7. We need to check if $d_4 = 13$ is consistent with the given conditions. If $d_4 = 13$, then:
$$d_m = (1 + d_2 + 13) d_8$$
Since $d_m = n$, we have:
$$n = (1 + d_2 + 13) d_8$$

8. We need to find $d_2$ and $d_8$ such that the equation holds. Let's assume $d_2 = 3$ (the smallest prime divisor of $n$). Then:
$$n = (1 + 3 + 13) d_8 = 17 d_8$$

9. To match the number of divisors, one prime must have a power of 2. Let's check if $n = 3^2 \cdot 13 \cdot 17$ fits the problem statement:
$$n = 9 \cdot 13 \cdot 17 = 1989$$

10. The divisors of 1989 are:
$$1, 3, 9, 13, 17, 39, 51, 117, 153, 221, 663, 1989$$
The 4th smallest divisor is 13, which matches $d_4 = 13$.

11. Therefore, $n = 1989$ satisfies all the given conditions.

The final answer is $\boxed{1989}$.