A representation of the Bernoulli numbers Let 
\begin{equation}
 \ell_{m,p}:=\sum_{j=1}^m\gamma_{m,j}\sigma_{p,j},
\end{equation}
where 
\begin{equation}
 \gamma_{m,j}:=\frac{2 (-1)^{j-1} }{j }\binom{2 m}{m+j}\Big/\binom{2 m}{m},\quad
 \sigma_{p,j}:=\sum _{i=0}^{j-1} \left(\frac{j}{2}-i-1\right)^p, 
\end{equation}
$p$ and $m$ are natural numbers, and 
\begin{equation}
m\ge m_p:=\left\lceil \frac{p+1}{2}\right\rceil.  
\end{equation}

Problem 1 (the more important one to me): Show that $\ell_{m,p}$ does not depend on $m$ (as long as $m\ge m_p$).



Problem 2 (a more specific and difficult (?) form of Problem 1): Show that (again for $m\ge m_p$)
  \begin{equation}  \ell_{m,p}=B_p,
 \tag{1} \end{equation} where $B_p$ is the $p$th Bernoulli number --
  see e.g.  \url{http://mathworld.wolfram.com/BernoulliNumber.html}
  and/or \url{https://en.wikipedia.org/wiki/Bernoulli_number}.  Formula
  $(1)$ has been verified for $p=1,\dots,20$ and $m=m_p,\dots,m_p+5$.

I think I know how to show that the restriction $m\ge m_p$ cannot be relaxed, for any given natural $p$.
These problems arise in a certain work in approximation theory. 
 A: I'll sketch the idea behind your claims, starting with Problem 1.
Step 1: Convince yourself that if we denote
$$F_p(j):=\frac1j\sum_{i=0}^{j-1}\left(\frac{j}2-i-1\right)^p$$ 
then $F_p(j)$ is always an even polynomial in $j$; a polynomial in $j^2$. For example, when $p$ is odd, we get $F_p(j)=-\left(\frac{j}2\right)^{p-1}$. 
Step 2: Therefore, from Step 1, it suffices to consider even powers of $j$ instead of $F_p(j)$. We shall also disregard $\frac{-2}{\binom{2m}m}$ and focus on the sum
$$G_p(m):=\sum_{j=1}^m(-1)^j\binom{2m}{m-j}j^{e}$$
where $e\geq0$ is a fixed even integer. Observe that 
$$\sum_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e
=\begin{cases} 2G_p(m) \qquad \qquad \text{if $e\neq0$} \\
  2G_p(m)+1 \qquad \,\,\text{if $e=0$}.\end{cases}$$
Step 3: After re-indexing $k=m-j$,
$$\sum_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e=(-1)^m\sum_{k=0}^{2m}(-1)^k\binom{2m}k(m-k)^e.$$
It's well-known (actually the crux of the matter here) that if $n>a$ then we've the vanishing of
$$\sum_{k=0}^n(-1)^k\binom{n}k k^a=0.\tag{0}$$
The reason why you get a persistent value (for $m\geq m_p$) is caused by the one single term when $e=0$ which happens for $p$ even.
At any rate, we gather that if $e$ is even and $m\geq \frac{e}2$ then
$$G_p(m)=\begin{cases} \,\,\,\,\,0 \qquad \,\text{when $e\neq0$} \\
-\frac12 \qquad \text{when $e=0$}.
\end{cases}$$
To understand Problem 2, it remains to identify the constant term in $F_p(j)$ w.r.t. the variable $j$. To this end, the Binomial Theorem furnishes
\begin{align} F_p(j)
=\frac1j\sum_{i=1}^j\sum_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-k}(-1)^ki^k
=\sum_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-1-k}(-1)^k\sum_{i=1}^ji^k
\tag{1}
\end{align}
revealing that the only way to encounter the sought-after constant term is provided that $k=p$. Keeping in mind that $\sum_{i=1}^ji^k$ is divisible by $j$, we just look at the quantity (using know expressions for Bernoulli numbers)
$$\frac{(-1)^p}j\sum_{i=1}^ji^p=(-1)^p\left[j^{p-1}+\frac1{p+1}\sum_{r=0}^p\binom{p+1}rB_rj^{p-r}\right].\tag{2}$$
Thus, the term we seek is $(-1)^pB_p=B_p$ (remember: $p$ is even) leading to
\begin{align}
\frac2{\binom{2m}m}\sum_{j=1}^m(-1)^{j-1}\binom{2m}{m+j}F_p(j)
&=\frac{2B_p}{\binom{2m}m}\sum_{j=1}^m(-1)^{j-1}\binom{2m}{m+j} \\
&=\frac{2B_p}{\binom{2m}m}\frac12\binom{2m}m \\
&=B_p
\end{align}
which is exactly what we want to arrive at. We've already known from above the $B_p=0$ when $p$ is odd. The proof is complete for both problems.
