$\newcommand{\R}{\mathbb R}\renewcommand{\le}{\leqslant}\renewcommand{\ge}{\geqslant}$Let $n:=m\ge2$.
We will show that the desired condition holds if and only if
$$p\ge\frac{\lceil n/2\rceil-1}{2 \lceil n/2\rceil-1}.$$
Suppose for a moment that there exist random variables $X_1,\dots,X_n$ such that $X_i\sim {\rm Ber}(1/2)$ for all $i$ and $P(X_i=X_j)=p\in(0,1)$ if $i\ne j$. That is, there is a nonnegative function $g\colon\{0,1\}^n\to\R$ such that
(i) $\sum_{x\in\{0,1\}^n}g(x)=1$,
(ii) $\sum_{x\in\{0,1\}^n}1(x_i=0)g(x)=\frac12$ for all $i$,
(iii) $\sum_{x\in\{0,1\}^n}1(x_i=x_j)g(x)=p$ if $i\ne j$.
Then, by symmetry, conditions (i)--(iii) will hold with $\tilde g(x):=\frac1{n!}\sum_{\pi\in S_n}g(\pi(x))$ in place of $g(x)$, where $S_n$ is the set of all permutations of the set $[n]:=\{1,\dots,n\}$. Note that $\tilde g(x)=f(\sum_1^n x_i)$ for some function $f\colon\{0,\dots,n\}\to\R$ and all $x=(x_1,\dots,x_n)\in\{0,1\}^n$. Moreover, the conditions (i)--(iii) can be rewritten as
(I) $\sum_{k=0}^n \binom nk f(k)=1$,
(II) $\sum_{k=0}^n \binom{n-1}k f(k)=\frac12$ for all $i$,
(III) $\sum_{k=0}^n a_{n,k} f(k)=p$,
where
\begin{equation}
a_{n,k}=\binom{n-2}k+\binom{n-2}{k-2};
\end{equation}
of course, $\binom{n-2}k=0$ if $k\ge n-1$ and $\binom{n-2}{k-2}=0$ if $k\le1$.
Thus, for any given $n\ge2$ and $p\in(0,1)$, we want to see whether there is a nonnegative function $f\colon\{0,\dots,n\}\to\R$ such that conditions (I)--(III) hold.
Consider the ratios
\begin{equation}
r_k:= r_{n,k}:=\frac{a_{n,k}}{\binom nk}
=\frac{(n-k)(n-k-1)+k (k-1)}{n (n-1)}.
\end{equation}
Note that $r_{k+1}\le r_k$ if $0\le k\le\frac{n-1}2$ and $r_{k+1}\ge r_k$ if $\frac{n-1}2\le k\le n-1$. Also, $r_k=r_{n-k}$.
It follows that
the maximum, say $\overline p_n$, of $\sum_{k=0}^n a_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that condition (I) holds is attained if $f(0)=f(n)=\frac12$ and $f_1=\cdots=f_{n-1}=0$, and hence $\overline p_n=1$; moreover, then condition (II) happens to hold as well;
if $n=2m$ is even, then the minimum, say $\underline p_n$, of $\sum_{k=0}^n a_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that condition (I) holds is attained if $f(m)=1/\binom nm=1/\binom{2m}m$ and $f_k=0$ for $k\in\{0,\dots,n\}\setminus\{m\}$, and hence $\underline p_n=\underline p_{2m}=\frac{m-1}{2m-1}$; moreover, then condition (II) happens to hold as well;
if $n=2m+1$ is odd, then the minimum $\underline p_n$ of $\sum_{k=0}^n a_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that condition (I) holds is attained if $f(m)=f(m+1)=\frac12/\binom{2m+1}m=\frac12/\binom{2m+1}{m+1}$ and $f_k=0$ for $k\in\{0,\dots,n\}\setminus\{m,m+1\}$, and hence $\underline p_n=\underline p_{2m+1}=\frac m{2m+1}$; moreover, then condition (II) happens to hold as well.
So,
the maximum of $\sum_{k=0}^n a_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that conditions (I) and (II) hold is $\overline p_n=1$;
if $n=2m$ is even, then the minimum of $\sum_{k=0}^n a_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that conditions (I) and (II) hold is $\underline p_n=\frac{m-1}{2m-1}$;
if $n=2m+1$ is odd, then the minimum of $\sum_{k=0}^n a_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that conditions (I) and (II) hold is $\underline p_n=\frac m{2m+1}$.
The set of all values of $\sum_{k=0}^n a_{n,k} f(k)$, where $f\colon\{0,\dots,n\}\to\R$ is a nonnegative function such that conditions (I) and (II) hold, is convex and therefore coincides with the interval $[\underline p_n,\overline p_n]=[\underline p_n,1]$.
Thus, the following two conditions are equivalent to each other:
there is a nonnegative function $f\colon\{0,\dots,n\}\to\R$ such that conditions (I)--(III) hold;
$\underline p_n\le p\le1$.
That is, for each natural $n\ge2$ and each $p\in[0,1]$, the following two conditions are equivalent to each other:
One may also note that
$\underline p_n=\dfrac{m_n-1}{2m_n-1}$, where $m_n:=\lceil n/2\rceil$;
$\underline p_n\le\underline p_{n+1}<\frac12$ for all $n\ge2$;
$\underline p_n\to\frac12$ as $n\to\infty$.
This equality does not hold in general. E.g., suppose that $n=2$,
$$\big(P_p(X_1=i,X_2=j)\colon\; i=0,1,\,j=0,1\big)=
\frac1{30}\,\begin{pmatrix}
4 & 9 \\
9 & 8 \\
\end{pmatrix},$$
and
$$\big(P_q(X_1=i,X_2=j)\colon\; i=0,1,\,j=0,1\big)=
\frac1{17}\,\begin{pmatrix}
2 & 6 \\
2 & 7 \\
\end{pmatrix}.$$
Then the left- and right-hand sides of your conjectured equality are $0.132\ldots$ and $0.118\ldots$, respectively (assuming $\log=\ln$).
Here are the calculations, done in Mathematica:
Best Answer
$\newcommand\X{\mathbf X}\newcommand\v{\mathbf v}$
Indeed, $1\le\sum_{i=1}^n X_i\le n$ and hence $$n\sum_{i=1}^n X_i\ge n=En\ge E\sum_{i=1}^n X_i,$$ so that $$\frac1n\,E\frac{\langle\X,\v\rangle}{\sum_{i=1}^n X_i} =E\frac{\langle\X,\v\rangle}{n\sum_{i=1}^n X_i}\le E\frac{\langle\X,\v\rangle}{E\sum_{i=1}^n X_i} =\frac{E\langle\X,\v\rangle}{E\sum_{i=1}^n X_i},$$ as claimed. $\quad\Box$