I claim the maximum is $n$. Notice that we can take $1,3,5,\dots,2n-1$. This clearly works. Now, we will show we cannot take more than $n$. Assume we can choose $n+1$ such integers. Then, there must clearly be at least one pair of consecutive integers $c$ and $c+1$. However, we have that $(c,c+1) = 1 = (c+1) - c$, a contradiction. Therefore, the maximum possible number of integers that one can choose is $n$, as desired.