The problem is directly based on this post, however, here the bound is lowered (it is not difficult to deal with the small cases, yet, they do require dedicated attention).

Answer: for n $\le$ 9 and odd n's wins $\boxed{Maris}$, for even n $\ge$ 10 wins $\boxed{Fillips}$. Let's look at 2 cases: 1) n is odd In the first move, Maris crosses out n, and creates a new game Fillips starting first and n being even. Now, let's divide numbers into $\frac{n-1}{2}$ groups: {1; 2}, {3; 4}, ..., {n-2; n-1}. Fillips starts first, so Maris can cross out groups from which Fillips selected a number. Then, uncrossed numbers would be consecutive and coprime. So, $\boxed{Maris}$wins. 2) n is even 2.1) n $\le$ 8 It's easy to see that for n=4, 6, 8, $\boxed{Maris}$ wins by crossing even numbers except 2. 2.2) n $\ge$ 10 There are $\frac{n}{2}$ even numbers, and Maris has $\frac{n-2}{2}$ moves. So if Fillips won't cross out even numbers, Maris can only cross out even numbers(2 even numbers aren't coprime) and cannot cross out odd numbers. Then, if Fillips will cross out only odd numbers except 3 and 9 in the first $\frac{n-4}{2}$ moves and then will cross out 1 even number left in the last move, $\boxed{Fillips}$ will win.