Let $\sigma_{1}(N)$ denote the sum of the divisors of the natural number $N$.
Here is my question:
For which natural numbers $N$ does $2N-\sigma(N)= D(N) \mid N$ hold?
$D(N)$ is the deficiency of $N$.
MY ATTEMPT
When $N=p$ is prime an odd prime, then
$$D(N)=D(p)=2p-\sigma(p)=2p-(p+1)=p-1$$
which does not divide $N=p$. (In fact, $\gcd(N, D(N))=\gcd(p,p-1)=1$.)
When $N=p^k$ is a prime power an odd prime power, then
$$D(N)=D(p^k)=2{p^k}-\sigma(p^k)=2{p^k}-\left(1 + p + \ldots + p^k\right)={p^k}-\sigma(p^{k-1}).$$
Now, since $\gcd(p,\sigma(p^{k-1}))=1$, then $\gcd(p^k,\sigma(p^{k-1}))=1$. Consequently, $$\gcd(N,D(N))=\gcd({p^k},{p^k}-\sigma(p^{k-1}))=1.$$
It follows that ${p^k}-\sigma(p^{k-1})=D(N) \nmid N$.
ADDED SEPTEMBER 26 2016
OOPS. When $N=2^r$ and $r \geq 0$, then $D(N)=2N-\sigma(N)=2^{r+1}-\frac{2^{r+1} - 1}{2 - 1}=1$. Thus, for $N=2^r$ and $r \geq 0$, $D(N) \mid N$.
Which leads me to ask:
Is $N=2^r$ for $r \geq 0$ the only (infinite) family of natural numbers satisfying $2N-\sigma(N)=D(N) \mid N$?
At this point, it seems hard to generalize to $\omega(N) \geq 2$, where $\omega(N)$ is the number of distinct prime factors of $N$.
If you are only considering numbers $n$ for which $2n-\sigma(n)>0$, they are tabulated here, where they are called "deficient-perfect numbers." There are plenty of numbers listed there that aren't powers of 2. It doesn't say whether there are infinitely many, but there are links to some references – maybe you could check them out, and report your findings back to us. It does say,
"If $2^{k+1} + 2^t - 1$ is an odd prime and $t \le k$, then $n = 2^k(2^{k+1} + 2^t - 1)$ is deficient-perfect with $2n - \sigma(n) = 2^t$. In fact, these are the only terms with two distinct prime factors."
If you allow abundant numbers, too, well, they're probably tabulated at OEIS as well. I didn't check.