By "discrete continuity" it suffices to find a set of $2014$ consecutive positive integers with $0$ special numbers and a different one with $2014$ special numbers. Then as we "slide" from one set to the other we are guaranteed to end up with some set with $2012$ special numbers at some point in the process, since this number changes by at most $1$ each time. For the set with $0$ special numbers, just note that the density of special numbers in $\mathbb{Z}^+$ is $0$. For the set with $2014$ special numbers, take the numbers starting with $2^{\mathrm{lcm}(2,\ldots,2015)}+2$.