Belarusian Mathematical Olympiad

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 $1001$.