Up! Up!

With not too much pain, we can directly compute the winning player for each n. Let's let $f(n) = 1$ or $2$, corresponding to the player order of the winning player given a starting number $n$ (so $f(1)=1$, for example). Note that $f(k) = 1$ iff $\exists(x<2016) : f(x)=2$, or $2014|x$ and $f(x/2014)=2$. Let's define $f(0) = 2$. It's not too hard to see that if $f(x)=2$, the next largest value of $k$ with $f(k)=2$ is $2016+x$ if $2014$ doesn't divide $2016+x$. So the values where Player 2 wins, up till $2014*2016$, are neatly the multiples of $2016$. The next value, however, is $2015^2$. Now, all the remaining values where Player 2 wins cannot be divisible by $2014$, since they're all odd. Thus, they form an AP of the form $2015^2+2016n$. The AP that Ozna picks must contain this AP entirely (else Navi wins for an infinite number of starting numbers), hence its common difference must divide $2016$ (hence it is $2016$). It's clear then that none of the earlier Player 2-winning numbers can be contained in this AP, so all $2013$ of them are available.