Combinatorial Identity involving Bernoulli numbers For all $n\geq 1$ and $m\geq0$, I'm trying to prove that
$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$
where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.
I made a couple of attempts using the recursive relationship of Bernoulli numbers and induction, but unsuccessful. Now I'm wondering if this is a viable proof strategy. Any comments are appreciated!
Edit: This is a conjecture, I evaluated it numerically for $n,m\leq 20$.

The instructive derivation of @RenéGy reduces OPs problem to show the nice symmetric Bernoulli number identity:
\begin{align*}
\color{blue}{(-1)^n \sum_{g=0}^m}&\color{blue}{ \binom{m}{g}\frac{B_{n+g+1}}{n+g+1}
+(-1)^m \sum_{g=0}^n \binom{n}{g}\frac{B_{m+g+1}}{m+g+1}}\\
&\color{blue}{= - \frac{1}{(n+m+1){\binom{n+m}{m}}}}\tag{1}
\end{align*}

Here we show the identity (1) follows from Theorem 2 in Bernoulli Numbers and a New Binomial Transform Identity by H.W. Gould.

Given a sequence $(a_n)_{n\geq 0}$ we consider the sequence of binomial transforms $(b_n)_{n\geq 0}$. We so have the binomial inverse pair
\begin{align*}
b_n=\sum_{k=0}^n\binom{n}{k}a_k\qquad\mathrm{and}\qquad a_n=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}b_k\tag{2}
\end{align*}
Setting $a_k=B_k$, the $k$-th Bernoulli number we have thanks to a well known recursion formula of the Bernoulli numbers the binomial inverse pair
\begin{align*}
a_k=B_k\quad\mathrm{and}\quad  b_n=\sum_{k=0}^n\binom{n}{k}a_k=\sum_{k=0}^n\binom{n}{k}B_k=(-1)^nB_n\qquad\qquad n\geq 0
\end{align*}
Theorem 2 in the paper states the following identity for binomial inverse pairs:
\begin{align*}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n+k+s}{s}}a_{n+k+s}
&=\sum_{k=0}^n(-1)^{n-k}\frac{\binom{n}{k}}{\binom{m+k+s}{s}}b_{m+k+s}\\
&\quad+\sum_{j=0}^{s-1}\sum_{i=0}^{s-1-j}\binom{s-1-j}{i}\binom{s-1}{j}\frac{(-1)^{n+1+i}sa_j}{(m+n+1+i)\binom{m+n+i}{n}}
\end{align*}
Setting $s=1$ the identity reduces to
\begin{align*}
\sum_{k=0}^m\binom{m}{k}\frac{a_{n+k+1}}{n+k+1}&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\frac{b_{m+k+1}}{m+k+1}\\
&\qquad+\frac{(-1)^{n+1}a_0}{(m+n+1)\binom{m+n}{n}}\tag{3}
\end{align*}

Putting the Bernoulli numbers (2) in (3) we obtain with $a_0=B_0=1$:
\begin{align*}
\sum_{k=0}^m\binom{m}{k}\frac{B_{n+k+1}}{n+k+1}
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\frac{(-1)^{m+k+1}B_{m+k+1}}{m+k+1}\\
&\qquad+\frac{(-1)^{n+1}}{(m+n+1)\binom{m+n}{n}}\\
\color{blue}{(-1)^n\sum_{k=0}^m\binom{m}{k}\frac{B_{n+k+1}}{n+k+1}}
&\color{blue}{=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{m+1}B_{m+k+1}}{m+k+1}}\\
&\qquad\color{blue}{-\frac{1}{(m+n+1)\binom{m+n}{n}}}
\end{align*}
and the claim (1) follows.

