Problem 1239

1. Let $d_1 = m$ and $d_2 - d_1 = a$. Then the divisors in arithmetic progression are $d_1, d_2, \ldots, d_k$ where $d_i = m + (i-1)a$ for $i = 1, 2, \ldots, k$.

2. The sum of these divisors is:
$$n = d_1 + d_2 + \cdots + d_k = m + (m + a) + (m + 2a) + \cdots + (m + (k-1)a)$$
This can be simplified using the formula for the sum of an arithmetic series:
$$n = k \cdot m + a \cdot \frac{k(k-1)}{2}$$

3. Since $d_k \mid n$, we have:
$$m + (k-1)a \mid k \cdot m + a \cdot \frac{k(k-1)}{2}$$

4. Let $\gcd(m, a) = d$. Then $m = dm_1$ and $a = da_1$ with $\gcd(m_1, a_1) = 1$. Substituting these into the divisibility condition, we get:
$$m_1 + (k-1)a_1 \mid m_1k + a_1 \cdot \frac{k(k-1)}{2}$$

5. Similarly, since $d_{k-1} \mid n$, we have:
$$m_1 + (k-2)a_1 \mid m_1k + a_1 \cdot \frac{k(k-1)}{2}$$

6. Using the Euclidean algorithm, we find:
$$\gcd(m_1 + (k-1)a_1, m_1 + (k-2)a_1) = \gcd(a_1, m_1 + (k-2)a_1) = \gcd(m_1, a_1) = 1$$

7. Therefore, from the above, we obtain:
$$(m_1 + (k-1)a_1) \cdot (m_1 + (k-2)a_1) \mid m_1k + a_1 \cdot \frac{k(k-1)}{2}$$

8. Expanding and simplifying, we get:
$$m_1^2 + m_1a_1(2k-3) + a_1^2(k-1)(k-2) \mid m_1k + a_1 \cdot \frac{k(k-1)}{2}$$

9. From this, we deduce that $2 \le k \le 3$.

Case 1: If $k = 2$:
$$m_1^2 + m_1a_1 \mid 2m_1 + a_1$$
Since $\gcd(m_1, a_1) = 1$ and $m_1 \mid 2m_1 + a_1$, we get $m_1 = 1$. This implies:
$$a_1 + 1 \mid 2 + a_1$$
which is a contradiction.

Case 2: If $k = 3$:
$$m_1^2 + 3m_1a_1 + 2a_1^2 \mid 3m_1 + 3a_1$$
From here, we get $a_1 = 1$. Therefore:
$$m_1^2 + 2m_1 + 2 \mid 3m_1 + 3$$
This implies $m_1 = 1$. Hence, $m = a$ and $k = 3$.

10. Therefore, the divisors are $d_1 = m, d_2 = 2m, d_3 = 3m$ and:
$$n = d_1 + d_2 + d_3 = 6m$$

Thus, the solution is $n = 6m$ for all $m \in \mathbb{Z}^+$.

The final answer is $\boxed{ n = 6m }$ for all $m \in \mathbb{Z}^+$.