Answer: $n = 924$.
Let $d_1 < d_2 < d_3$ be the divisors given in the condition, then $1 < d_1 < d_2 < d_3 < n$. Since $d_1$, $d_2$ and $d_3$ are divisors of $n$, there exist positive integers $a$, $b$ and $c$ such that $d_1 a = d_2 b = d_3 c = n$. It is clear that $1 < c < b < a < n$. Hence $c \ge 2$, $b \ge 3$, $a \ge 4$. The equation $d_1 + d_2 + d_3 = 1001$ can be transformed:
$$1001 = \frac{n}{a} + \frac{n}{b} + \frac{n}{c} = n \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \le n \left( \frac{1}{4} + \frac{1}{3} + \frac{1}{2} \right) = \frac{13n}{12},$$
so $12 \cdot 1001 \le 13n$, i.e. $12 \cdot 77 \le n$, $n \ge 924$. On the other hand, the value $n = 924$ satisfies the condition of the problem, since the number $924$ has proper divisors $462$, $308$ and $231$ the sum of which $462 + 308 + 231$ equals to $1001$.