Is there a deficient $N$ such that $\sqrt{D(N)} \mid A(N)$ holds, where $D(N)$ is the deficiency and $A(N)$ is the sum of the aliquot parts of $N$?

Question

Do there exist deficient numbers $N$ such that $$\sqrt{D(N)} \mid A(N)$$ holds, where $$D(N) = 2N - \sigma(N)$$ is the deficiency and $$A(N) = \sigma(N) - N$$ is the sum of aliquot parts of $N$?