Your numbers have positive lower density. To see this let $z$ be a positive integer to be fixed later, and denote $$ c:=\prod_{p \leq z}(1-1/p). $$ Consider all square-free integers $x < n \leq 2x$ which are composed of primes $z < p \leq \sqrt{x}$. Note that these numbers satisfy the requirements. Their number, by a crude estimate, is at least $$ cx+O(1)-\sum_{\sqrt{x} < p \leq 2x}(cx/p+O(1)) - \sum_{p>z}x/p^2 sum_{z < p \geq (c-c\log leq \sqrt{2x}}(cx/p^2 +O(1)), $$ which is at least $$ c(1-\log 2-1/z+o(1))x. $$ That is, we have positive the lower density once is at least $$ c(1-\log 2) > 1/z. $2-1/z)/2.$$ For $The z:=5$ the left hand side is exceeds $\gg 1/\log (z)$, therefore the inequality holds 0.0142$, while for $z$ sufficiently large. Numerical calculation shows that $z:=23$ does the job, this choice proves that the lower density z:=17$ it exceeds 0.003$0.0223$.
EDIT: I improved slightly my original argument.
