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The answer is negative. Here is a counterexample. For natural $j$, let $$k_j:=\lfloor2^j/j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then $\frac1j-\frac1{2^j}<r_j\le\frac1j$ and hence $r_j$ is decreasing in $j\ge5$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$ Here $j\ge6$ is a natural number. Note that the intervals $\Delta_j$ are pairwise disjoint. Let then $$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$ for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and $$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$ Note that $\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$. Let, finally,
\begin{equation} B:=\bigcup_{j=6}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}. \end{equation}

Take now any large enough natural $n$. Let \begin{equation} H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big] \end{equation} for $i=0,\dots,2^n-1$.

From now on, suppose that $j\in\{n,\dots,2n\}$.

Note also that $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, given the condition $j\in\{n,\dots,2n\}$, where $|\cdot|$ denotes the Lebesgue measure. For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have $$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q} \quad\text{and}\quad H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$ for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$. Hence, $|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.
So, \begin{equation} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) =\frac14 \end{equation} -- for each $j\in\{n,\dots,2n\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.
Note here that the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is $\asymp\frac1{j^2}$, and so, for a given natural $n$, the dyadic interval $\Delta_j$ contains at least $c \frac{2^n}{j^2}$ dyadic intervals $H_{n,i}$, for some universal real constant $c>0$. So, for $a_n$ as defined in the question, we have
\begin{equation} a_n\ge\frac1{2^n}\sum_{j=n}^{2n}\ \sum_{i\colon H_{n,i}\subseteq\Delta_j} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) \ge \frac1{2^n}\sum_{j=n}^{2n}c\frac{2^n}{j^2}\frac14\asymp\frac1n, \end{equation} which yields $\sum_n a_n=\infty$.

Starting with $k_j:=\lfloor2^j\epsilon_j\rfloor$ instead of $k_j:=\lfloor2^j/j\rfloor$, where $(\epsilon_j)$ is any sequence of positive numbers converging to $0$ slowly and regularly enough, we can similarly conclude that $a_n$ can go to $0$ however slowly.

The answer is negative. In fact, let us show that $a_n$ as defined in the question may converge to $0$ however slowly. Indeed, let $(\epsilon_j)$ is any sequence in $(0,1]$ converging to $0$ slowly and regularly enough, in the sense that \begin{equation} \epsilon_{j-1}-\epsilon_j\ge2^{2-j}\tag{1} \end{equation} eventually in $j$ -- that is, for some natural $j_0$ and all natural $j\ge j_0$. For instance, this condition will hold if $\epsilon_j$ is (eventually in $j$) of the form $2^{2-j}$ or $j^{-a}$ or $(\underbrace{\ln\dots\ln}_N j)^{-a}$ for some real $a$ and some natural $N$.

In what follows, $j$ is a natural number $\ge j_0$.
Let $$k_j:=\lfloor2^j\epsilon_j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then \begin{equation} \epsilon_j-\tfrac1{2^j}<r_j\le\epsilon_j\tag{2} \end{equation} and hence, by the regularity condition $(1)$, $r_j<r_{j-1}$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$ Note that the intervals $\Delta_j$ are pairwise disjoint. Let then $$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$ for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and $$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$ Note that $\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$. Let, finally,
\begin{equation} B:=\bigcup_{j=j_0}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}. \end{equation}

Take now any large enough natural $n$. Let \begin{equation} H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big] \end{equation} for $i=0,\dots,2^n-1$.

From now on, suppose also that $j\ge n\ge j_0$.

Then $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, where $|\cdot|$ denotes the Lebesgue measure. For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have $$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q} \quad\text{and}\quad H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$ for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$. Hence, $|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.
So, \begin{equation} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) =\frac14 \end{equation} -- for each $j\in\{n,n+1,\dots\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.
By $(2)$ and $(1)$, the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is $r_{j-1}-r_j\ge\epsilon_{j-1}-\epsilon_j-\tfrac1{2^{j-1}}\ge\tfrac12\,(\epsilon_{j-1}-\epsilon_j)$, and so, for a given natural $n\ge j_0$, the dyadic interval $\Delta_j$ contains at least $2^{n-1}(\epsilon_{j-1}-\epsilon_j)$ dyadic intervals $H_{n,i}$. So, for $a_n$ as defined in the question, we have
\begin{equation} a_n\ge\frac1{2^n}\sum_{j=n}^\infty\ \sum_{i\colon H_{n,i}\subseteq\Delta_j} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) \ge \frac1{2^n}\sum_{j=n}^\infty 2^{n-1}(\epsilon_{j-1}-\epsilon_j)\frac14 =\frac{\epsilon_{n-1}}8. \end{equation} Thus, $a_n\ge\frac{\epsilon_{n-1}}8$ for all $n\ge j_0$. For instance, letting $\epsilon_j=1/j$ for all natural $j$, we have
$\sum_n a_n=\infty$.

The answer is negative. Here is a counterexample. For natural $j$, let $$k_j:=\lfloor2^j/j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then $\frac1j-\frac1{2^j}<r_j\le\frac1j$ and hence $r_j$ is decreasing in $j\ge5$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$ Here $j\ge6$ is a natural number. Note that the intervals $\Delta_j$ are pairwise disjoint. Let then $$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$ for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and $$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$ Note that $\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$. Let, finally,
\begin{equation} B:=\bigcup_{j=6}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}. \end{equation}

Take now a natural $n$. Let \begin{equation} H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big] \end{equation} for $i=0,\dots,2^n-1$. From now on, suppose that $j\in\{n,\dots,2n\}$.

Note also that $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, given the condition $j\in\{n,\dots,2n\}$, where $|\cdot|$ denotes the Lebesgue measure. For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have $$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q} \quad\text{and}\quad H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$ for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$. Hence, $|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.
So, \begin{equation} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) =\frac14 \end{equation} -- for each $j\in\{n,\dots,2n\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.
Note here that the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is $\asymp\frac1{j^2}$, and so, for a given natural $n$, the dyadic interval $\Delta_j$ contains at least $c \frac{2^n}{j^2}$ dyadic intervals $H_{n,i}$, for some universal real constant $c>0$. So, for all large enough $n$ and for $a_n$ as defined in the question, we have
\begin{equation} a_n\ge\frac1{2^n}\sum_{j=n}^{2n}\ \sum_{i\colon H_{n,i}\subseteq\Delta_j} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) \ge \frac1{2^n}\sum_{j=n}^{2n}c\frac{2^n}{j^2}\frac14\asymp\frac1n, \end{equation} which yields $\sum_n a_n=\infty$.

Starting with $k_j:=\lfloor2^j\epsilon_j\rfloor$ instead of $k_j:=\lfloor2^j/j\rfloor$, where $(\epsilon_j)$ is any sequence of positive numbers converging to $0$ slowly and regularly enough, we can similarly conclude that $a_n$ can go to $0$ however slowly.

The answer is negative. Here is a counterexample. For natural $j$, let $$k_j:=\lfloor2^j/j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then $\frac1j-\frac1{2^j}<r_j\le\frac1j$ and hence $r_j$ is decreasing in $j\ge5$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$ Here $j\ge6$ is a natural number. Note that the intervals $\Delta_j$ are pairwise disjoint. Let then $$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$ for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and $$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$ Note that $\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$. Let, finally,
\begin{equation} B:=\bigcup_{j=6}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}. \end{equation}

Take now any large enough natural $n$. Let \begin{equation} H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big] \end{equation} for $i=0,\dots,2^n-1$.

From now on, suppose that $j\in\{n,\dots,2n\}$.

Note also that $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, given the condition $j\in\{n,\dots,2n\}$, where $|\cdot|$ denotes the Lebesgue measure. For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have $$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q} \quad\text{and}\quad H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$ for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$. Hence, $|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.
So, \begin{equation} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) =\frac14 \end{equation} -- for each $j\in\{n,\dots,2n\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.
Note here that the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is $\asymp\frac1{j^2}$, and so, for a given natural $n$, the dyadic interval $\Delta_j$ contains at least $c \frac{2^n}{j^2}$ dyadic intervals $H_{n,i}$, for some universal real constant $c>0$. So, for $a_n$ as defined in the question, we have
\begin{equation} a_n\ge\frac1{2^n}\sum_{j=n}^{2n}\ \sum_{i\colon H_{n,i}\subseteq\Delta_j} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) \ge \frac1{2^n}\sum_{j=n}^{2n}c\frac{2^n}{j^2}\frac14\asymp\frac1n, \end{equation} which yields $\sum_n a_n=\infty$.

Starting with $k_j:=\lfloor2^j\epsilon_j\rfloor$ instead of $k_j:=\lfloor2^j/j\rfloor$, where $(\epsilon_j)$ is any sequence of positive numbers converging to $0$ slowly and regularly enough, we can similarly conclude that $a_n$ can go to $0$ however slowly.

The answer is negative. Here is a counterexample. For natural $j$, let $$k_j:=\lfloor2^j/j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then $\frac1j-\frac1{2^j}<r_j\le\frac1j$ and hence $r_j$ is decreasing in $j\ge5$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$ Here $j\ge6$ is a natural number. Note that the intervals $\Delta_j$ are pairwise disjoint. Let then $$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$ for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and $$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$ Note that $\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$. Let, finally,
\begin{equation} B:=\bigcup_{j=6}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}. \end{equation}

Take now a natural $n$. Let \begin{equation} H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big] \end{equation} for $i=0,\dots,2^n-1$. From now on, suppose that $j\in\{n,\dots,2n\}$.

Note also that $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, given the condition $j\in\{n,\dots,2n\}$. For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have $$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q} \quad\text{and}\quad H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$ for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$. Hence, for the Lebesgue measure $|B\cap H_{n,i}|$ of the set $B\cap H_{n,i}$ we have $|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.
So, \begin{equation} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) =\frac14 \end{equation} -- for each $j\in\{n,\dots,2n\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.
Note here that the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is $\asymp\frac1{j^2}$, and so, for a given natural $n$, the dyadic interval $\Delta_j$ contains at least $c \frac{2^n}{j^2}$ dyadic intervals $H_{n,i}$, for some universal real constant $c>0$. So, for all large enough $n$ and for $a_n$ as defined in the question, we have
\begin{equation} a_n\ge\frac1{2^n}\sum_{j=n}^{2n}\ \sum_{i\colon H_{n,i}\subseteq\Delta_j} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) \ge \frac1{2^n}\sum_{j=n}^{2n}c\frac{2^n}{j^2}\frac14\asymp\frac1n, \end{equation} which yields $\sum_n a_n=\infty$.

Starting with $k_j:=\lfloor2^j\epsilon_j\rfloor$ instead of $k_j:=\lfloor2^j/j\rfloor$, where $(\epsilon_j)$ is any sequence of positive numbers converging to $0$ slowly and regularly enough, we can similarly conclude that $a_n$ can go to $0$ however slowly.

The answer is negative. Here is a counterexample. For natural $j$, let $$k_j:=\lfloor2^j/j\rfloor$$ and $$r_j:=k_j/2^j.$$ Then $\frac1j-\frac1{2^j}<r_j\le\frac1j$ and hence $r_j$ is decreasing in $j\ge5$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$ Here $j\ge6$ is a natural number. Note that the intervals $\Delta_j$ are pairwise disjoint. Let then $$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$ for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and $$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$ Note that $\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$. Let, finally,
\begin{equation} B:=\bigcup_{j=6}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}. \end{equation}

Take now a natural $n$. Let \begin{equation} H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big] \end{equation} for $i=0,\dots,2^n-1$. From now on, suppose that $j\in\{n,\dots,2n\}$.

Note also that $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, given the condition $j\in\{n,\dots,2n\}$, where $|\cdot|$ denotes the Lebesgue measure. For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have $$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q} \quad\text{and}\quad H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$ for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$. Hence, $|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.
So, \begin{equation} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) =\frac14 \end{equation} -- for each $j\in\{n,\dots,2n\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.
Note here that the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is $\asymp\frac1{j^2}$, and so, for a given natural $n$, the dyadic interval $\Delta_j$ contains at least $c \frac{2^n}{j^2}$ dyadic intervals $H_{n,i}$, for some universal real constant $c>0$. So, for all large enough $n$ and for $a_n$ as defined in the question, we have
\begin{equation} a_n\ge\frac1{2^n}\sum_{j=n}^{2n}\ \sum_{i\colon H_{n,i}\subseteq\Delta_j} \frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big) \ge \frac1{2^n}\sum_{j=n}^{2n}c\frac{2^n}{j^2}\frac14\asymp\frac1n, \end{equation} which yields $\sum_n a_n=\infty$.

Starting with $k_j:=\lfloor2^j\epsilon_j\rfloor$ instead of $k_j:=\lfloor2^j/j\rfloor$, where $(\epsilon_j)$ is any sequence of positive numbers converging to $0$ slowly and regularly enough, we can similarly conclude that $a_n$ can go to $0$ however slowly.