This question was edited, in particular equations were corrected:
A number N is said to be deficient by an integer $d$ if:
$\sigma(N)=2N-d$
Note that powers of 2 are deficient by 1.
While a prime $p$ is deficient by $p-1$
My question will be this:
Is there an integer $N$ that is deficient by an integer $d$ that lies between $(2/3)N$ and $N-\sqrt{N}$, that is
$(2/3)N<d<N-\sqrt{N}$ and can we characterize them?
Thanks a lot for your help.
Since $\sigma(n)>0$ we have $d<2N$, together with $(2/3)N^2<d$ we get $(2/3)N^2<2N$ which is verified only for $0< N< 3$, which leaves only 2 cases to be checked, namely $1$ and $2$.
edit:After the question being asked has been changed the first part of my answer is irrelevant. I can't completely answer the new question, but I can show some solutions.
Consider a pair of twin primes $p$ and $p+2$ and their product $N=p^2+2p$.
$\sigma(N)$ is clearly given by $1+p+p+2+p^2+2p$, that is $\sigma(N)=p^2+4p+2$.
From $\sigma(N)=2N-d$ we get $d=2(p^2+2p)-(p^2+4p+2)=p^2-3$
Substituting into the inequality $(2/3)N<d<N-\sqrt{N}$ we get $$\frac{2}{3}(p^2+2p)<p^2-3<p^2+2p-\sqrt{p^2+2p}$$ The left inequality is satisfied for $p>2+\sqrt{13}$ while the right one is satisfied for $p\ge0$.
So as long as $p$ and $p+2$ are both prime and $p\ge11$ their product is a number with the required deficiency (whether there are infinite pairs of twin primes is an exercise left to the reader :P).
For example consider $p=29$, that gives $N=899$, $\sigma(N)=960$, $(2/3)N=599.33...$, $d=838$ and $N-\sqrt{N}=869.01...$
An analogous reasoning shows that if $p$ and $p+4$ are both primes and $p\ge7$ then their product is another solution, for example $77=7\times11$ and $221=13\times 17$. I suspect this can be generalized to more semiprimes, maybe all of them bigger than a certain number?