I've got some partial results.
This answer proves the following claims :
Claim 1 : If $N=2^k$, then $\frac{N}{D(N)}$ is an almost perfect number.
Claim 2 : If $N = 2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime and $0\lt t\le k$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Claim 3 : If $N$ is not of the form $2^k$ and $N\le 49024$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Claim 4 : If $\frac{N}{D(N)}=2^k$ where $k\ge 1$ with $D(N)\not=1$, then
$D(N)$ is even.
$D(N)$ has at least three distinct prime factors.
$D(N)$ is not of the form $2pq$ where $p,q$ are distinct odd primes.
$D(N)$ is not of the form $2^2pq$ where $p,q$ are distinct odd primes such that $(p,q)\not\equiv (1,1)\pmod 4$.
$D(N)\ge 90$
Claim 1 : If $N=2^k$, then $\frac{N}{D(N)}$ is an almost perfect number.
Proof : We have $D(N)=2^{k+1}-(2^{k+1}-1)=1$ and
$$\sigma\bigg(\frac{N}{D(N)}\bigg)-\bigg(\frac{2N}{D(N)}-1\bigg)=\sigma(2^k)-\sigma(2^k)=0\qquad\blacksquare$$
Claim 2 : If $N = 2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime and $0\lt t\le k$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Proof : If $N = 2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime, then we get
$$D(N)=2^{k+1}(2^{k+1} + 2^t - 1)-(2^{k+1}-1)(2^{k+1} + 2^t)=2^t$$
from which we have
$$\begin{align}&\sigma\bigg(\frac{N}{D(N)}\bigg)-\bigg(\frac{2N}{D(N)}-1\bigg)
\\\\&=\sigma\bigg(2^{k-t}(2^{k+1} + 2^t - 1)\bigg)-(2^{k+1}-1)(2^{k-t+1}+1)
\\\\&=(2^{k-t+1}-1)(2^{k+1}+2^t)-(2^{k+1}-1)(2^{k-t+1}+1)
\\\\&=(2^{k-t+1}+1)(1-2^t)
\\\\&\not=0\qquad\blacksquare\end{align}$$
Claim 3 : If $N$ is not of the form $2^k$ and $N\le 49024$, then $\frac{N}{D(N)}$ is not an almost perfect number.
Proof : In the following, red numbers are of the form $2^k$, and blue numbers are of the form $2^k(2^{k+1} + 2^t - 1)$ where $2^{k+1} + 2^t - 1$ is an odd prime and $t\le k$ where $a_n$ is the $n$-th deficient-perfect number and $$f(N)=\sigma\bigg(\frac{N}{D(N)}\bigg)-\bigg(\frac{2N}{D(N)}-1\bigg)$$Note here that $$\text{$\dfrac{N}{D(N)}$ is an almost perfect number}\iff f(N)=0$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\ \hline
a_n&\color{red}1&\color{red}2&\color{red}4&\color{red}8&\color{blue}{10}&\color{red}{16}&\color{red}{32}&\color{blue}{44}&\color{red}{64}&\color{red}{128}&\color{blue}{136}&\color{blue}{152}&\color{blue}{184}&\color{red}{256}&\color{red}{512}\\ \hline
f(a_n)&0&0&0&0&\not=0&0&0&\not=0&0&0&\not=0&\not=0&\not=0&0&0 \\\hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
n&16&17&18&19&20&21&22&23&24&25&26\\ \hline
a_n&\color{blue}{752}&884&\color{red}{1024}&\color{red}{2048}&\color{blue}{2144}&\color{blue}{2272}&\color{blue}{2528}&\color{red}{4096}&\color{red}{8192}&\color{blue}{8384}&\color{blue}{12224}
\\ \hline
f(a_n)&\not=0&-189&0&0&\not=0&\not=0&\not=0&0&0&\not=0&\not=0 \\\hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
n&27&28&29&30&31&32&33&34&35\\ \hline
a_n&\color{red}{16384}&17176&18632&18904&\color{red}{32768}&\color{blue}{32896}&\color{blue}{33664}&\color{blue}{34688}&\color{blue}{49024}
\\ \hline
f(a_n)&0&-111&-1863&-2205&0&\not=0&\not=0&\not=0&\not=0 \\\hline
\end{array}$$
Claim 4 : If $\frac{N}{D(N)}=2^k$ where $k\ge 1$ with $D(N)\not=1$, then
$D(N)$ is even.
$D(N)$ has at least three distinct prime factors.
$D(N)$ is not of the form $2pq$ where $p,q$ are distinct odd primes.
$D(N)$ is not of the form $2^2pq$ where $p,q$ are distinct odd primes such that $(p,q)\not\equiv (1,1)\pmod 4$.
$D(N)\ge 90$
Proof : Let $D(N)=m$. Then, we have
$$\begin{cases}2N-\sigma(N)=m
\\\\\frac{N}{2N-\sigma(N)}=2^k\end{cases}\implies\sigma(2^km)=m(2^{k+1}-1)\tag1$$
for some $k\ge 1$.
Suppose that $m\ge 3$ is odd. Then, we have
$$(1)\implies\sigma(2^k)\sigma(m)=m(2^{k+1}-1)\implies \sigma(m)=m$$
which is impossible.
Suppose that $m=2^s$ where $s\ge 1$. Then, we have
$$(1)\implies 2^{k+s+1}-1=2^s(2^{k+1}-1)$$which is impossible since LHS is odd while RHS isn't.
Suppose that $m=2^st^u$ where $s\ge 1$ and $t$ is an odd prime with $u\ge 1$. Then, we have
$$(1)\implies (2^{k+s+1}-1)(1+t+\cdots +t^u)=2^st^u(2^{k+1}-1)$$
There exists a positive integer $v$ such that $$1+t+\cdots +t^u=2^sv\tag2$$ to have
$$(2^{k+s+1}-1)v=t^u(2^{k+1}-1)$$
There eixsts a positive integer $w$ such that $$2^{k+1}-1=vw\tag3$$ to have
$$2^{k+s+1}=t^uw+1\tag4$$
Multiplying the both sides of $(3)$ by $2^s$ and using $(2)(4)$ give
$$1-2^s=(1+t+\cdots +t^{u-1})w$$
which is impossible since LHS is negative while RHS isn't.
So, we see that $D(N)$ has at least three distinct prime factors.
Suppose that $m=2pq$ where $p,q$ are distinct odd primes. Then, we get
$$(1)\implies (2^{k+1}-1)(p+1)(q+1)=2pq(2^{k+1}-1)$$
which is impossible since LHS is divisible by $4$ while RHS isn't.
Suppose that $m=2^2pq$ where $p,q$ are distinct odd primes and $(p,q)\not\equiv (1,1)\pmod 4$. Then, we get
$$(1)\implies (2^{k+3}-1)\cdot\frac{p+1}{2}\cdot\frac{q+1}{2}=pq(2^{k+1}-1)$$
which is impossible since LHS is even while RHS isn't.
Hence, it follows that $D(N)\ge 2\cdot 3^2\cdot 5=90$. $\quad\blacksquare$
Too long to comment :
Let $c:=\dfrac{\ln(4/3)}{\ln(13/9)}$. Using $(I(m^2))^{c} < I(m)$, one can get
$$m\le \dfrac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\implies D(m)\lt D(p^k)\tag1$$
which is better than $$m < p^k \implies D(m) < D(p^k).$$
Proof :
$(I(m^2))^c\lt I(m)$ is equivalent to
$$\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c\lt\frac{\sigma(m)}{m}\iff \sigma(m)\gt m\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c=m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$
Using $\sigma(m)\gt m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$, we get
$$D(p^k)-D(m)=D(p^k)-2m+\sigma(m)\gt D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$
So, we can say that if
$$D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c\ge 0\tag2$$
then $D(m)\lt D(p^k)$ where note that
$$(2)\iff m\le \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^c}$$ So, we can say that $(1)$ holds.
To see that $(1)$ is better than $m < p^k \implies D(m) < D(p^k)$, it is sufficient to prove that
$$p^k\lt \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\tag3$$
We have
$$\begin{align}(3)&\iff p^k\lt \frac{p^{k+1}-2p^k+1}{(p-1)\bigg(2-\bigg(\dfrac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)}
\\\\&\iff p^k(p-1)\bigg(2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)\lt p^{k+1}-2p^k+1
\\\\&\iff 2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\lt \frac{p^{k+1}-2p^k+1}{p^k(p-1)}
\\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt 2-\frac{p^{k+1}-2p^k+1}{p^k(p-1)}
\\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt \frac{p^{k+1}-1}{p^k(p-1)}
\\\\&\iff \frac{2p^k(p-1)}{p^{k+1}-1}\gt \bigg(\frac{p^{k+1}-1}{p^k(p-1)}\bigg)^{1/c}
\\\\&\iff 2\gt \bigg(\frac{p^{k+1}-1}{p^{k}(p-1)}\bigg)^{(c+1)/c}
\\\\&\iff 2^{c/(c+1)}\gt \frac{p^{k+1}-1}{p^{k}(p-1)}
\\\\&\iff 2^{c/(c+1)}p^k(p-1)-p^{k+1}+1\gt 0
\\\\&\iff p^k\underbrace{\bigg((2^{c/(c+1)}-1)p-2^{c/(c+1)}\bigg)}_{\text{positive}}+1\gt 0\end{align}$$
which does hold. So, $(3)$ holds.
Best Answer
On OP's request, I am converting my comment into an answer.
The implication says only that if $a\le b$ and $c\le d$, then $ac\le bd$ (assuming that $a,c$ are non-negative).
It does not say that $ac\le bd$ only if $a\le b$ and $c\le d$.
(It is false that if $ac\le bd$, then $a\le b$ and $c\le d$.
For $(a,b,c,d)=(3,2,2,4)$, for example, we have $ac\le bd$ and $a\gt b$.)