8. The answer is (B). Consider a sequence $b_{1}, b_{2}, \ldots, b_{2017}$ of distinct positive integers whose sum is 2016$\cdot$2017$\cdot$2018. Let $d$ be the greatest common divisor of the elements of this sequence. By definition, the numbers $a_{1}:=b_{1} / d, a_{2}:=b_{2} / d, \ldots, a_{2017}:=b_{2017} / d$ are also distinct positive integers, and their sum is at least the sum $1+2+\cdots+2017$. We then have

$2016 \cdot 2017 \cdot 2018=b_{1}+\cdots+b_{2017}=d \cdot\left(a_{1}+a_{2}+\cdots+a_{2017}\right) \geq d \cdot(1+2+\cdots+2017)=d \cdot \frac{2017 \cdot 2018}{2}$,

from which $d \leq 2 \cdot 2016$. On the other hand, if we let $D=2 \cdot 2016$, the sequence $D, 2 D, 3 D, \cdots, 2017 D$ is one of the sequences that are written on the board, and clearly the greatest common divisor of its elements is precisely $D$.
8. Luca writes on a blackboard all possible sequences consisting of 2017 distinct positive integers whose sum is $2016 \cdot 2017 \cdot 2018$. After doing this, he replaces each of these sequences with the greatest common divisor of its elements. When this long operation is completed, what is the maximum of the numbers written on the blackboard?
(A) 2
(B) $2 \cdot 2016$
(C) $2 \cdot 2017$
(D) $2 \cdot 2018$
(E) $2016 \cdot 2018$