Amelie can force a victory. For the first number, she chooses 2017. After that, there are 252 numbers 1 mod 8, 2 mod 8, 3 mod 8, 4 mod 8... left. Then, whatever number Boris chooses, Amelie chooses the complement of Boris's number (Where the C(1) = 7, C(2) = 6, C(3) = 5, C(4) = 4, and C(8) = 8.) Since there are an even number of elements for each complement, Amelie can use this strategy until the last two numbers. Furthermore, the last two numbers will be complements of each other, and their sum will be divisible by each other.

I think Amelie wins. First she removes $2017$ and now the "new game" is given the numbers $1,2,3,\dots,2016$, Boris starts and the winning conditions are the same. Divide the 2016 numbers into groups of 8 of the form $\{8k-7, 8k-6, \dots, 8k\}$ for each $k= \overline{1,252}$. Now for each number $b$ Boris chooses, Amelie chooses the number $a$ such that $a \equiv -b \mod 8$ and both $a$ and $b$ are inside the same group. Notice that Amelie can do this whenever $b$ is different from $0,4 \mod 8$. In that case Amelie can just choose a number with the same properties as above, however in that case from a different group. Since the number of groups is even, Amelie can always do this. Side noteNote that the group part is totally unnecessary, however it makes the solution more understandable I guess, and kind of gives a "more precise" strategy.