An identity involving binomial coefficients and Bernoulli numbers. While solving a problem I came across the following identity, which holds by numerical evidence:
$$
\sum_{k=1}^i\frac1k\binom{i}{k-1}\binom kj{B_{k-j}}=\delta_{ij}.
$$
where $B$ are the Bernoulli numbers.
I have no experience with Bernoulli numbers, so any hint for proving the equality will be appreciated.
 A: We have the following claim where $n\ge j$ (the sum is zero when $n\lt
j$ and the claim holds by inspection)
$$\sum_{k=j}^n
\frac{1}{k} {n\choose k-1} {k\choose j} B _{k-j}
= \delta_{nj}.$$
This is
$$\sum_{k=j}^n
{n+1\choose k} {k\choose j} B _{k-j}
= \delta_{nj}  \times (n+1).$$
Now
$${n+1\choose k} {k\choose j} =
\frac{(n+1)!}{(n+1-k)! \times j! \times (k-j)!}
= {n+1\choose j} {n+1-j\choose k-j}$$
and we find
$$\sum_{k=j}^n
{n+1-j\choose k-j} B_{k-j}
= \delta_{nj}  \times (n+1) \times {n+1\choose j}^{-1}$$
or
$$\sum_{k=0}^{n-j}
{n+1-j\choose k} B_{k}
= \delta_{nj}  \times (n+1) \times {n+1\choose j}^{-1}
\\ = \delta_{nj}  \times (n+1) \times {n+1\choose n}^{-1}
= \delta_{nj}.$$
To prove this last form we put on the LHS
$$-B_{n+1-j} + \sum_{k=0}^{n+1-j}
{n+1-j\choose k} B_{k}
\\ = -B_{n+1-j}
+ (n+1-j)! [z^{n+1-j}] \frac{z}{\exp(z)-1} \exp(z).$$
Observe that
$$\frac{z}{\exp(z)-1} \exp(z) =
\frac{z}{\exp(z)-1} (\exp(z)-1) +
\frac{z}{\exp(z)-1}
\\ = z + \frac{z}{\exp(z)-1}$$
so that we get
$$-B_{n+1-j}
+ (n+1-j)! [z^{n+1-j}] z
+ (n+1-j)! [z^{n+1-j}] \frac{z}{\exp(z)-1}
\\ = - B_{n+1-j}
+ (n+1-j)! \delta_{nj}
+ B_{n+1-j}
= \delta_{nj},$$
which is the RHS. This concludes the argument.
A: Allow me to change the notation as to reserve (as far as possible) to $i$ it's common meaning,
and use $k,j,l$ as indices.
First let's simplify the sum working on the properties of the binomial coefficients
$$
\eqalign{
  & S(n,m) = \sum\limits_{1 \le \,k\, \le \,n} {{1 \over k}\left( \matrix{
  n \cr 
  k - 1 \cr}  \right)\left( \matrix{
  k \cr 
  m \cr}  \right)B_{\,k - m} }  =   \cr 
  &  = {1 \over {n + 1}}\sum\limits_{1 \le \,k\, \le \,n} {\left( \matrix{
  n + 1 \cr 
  k \cr}  \right)\left( \matrix{
  k \cr 
  m \cr}  \right)B_{\,k - m} }  =   \cr 
  &  = {1 \over {n + 1}}\left( \matrix{
  n + 1 \cr 
  m \cr}  \right)\sum\limits_{1 \le \,k\, \le \,n} {\left( \matrix{
  n + 1 - m \cr 
  k - m \cr}  \right)B_{\,k - m} }  =   \cr 
  &  = {1 \over {n + 1}}\left( \matrix{
  n + 1 \cr 
  m \cr}  \right)\sum\limits_{\max \left( {1 - m,0} \right) \le \,j\, \le \,n - m} {\left( \matrix{
  n + 1 - m \cr 
  j \cr}  \right)B_{\,j} }  \cr} 
$$
where:
 - in the first step, we use the "absorption" indentity;
 - in the second step, we use the "trinomial revision" indentity;
 - in the third step, we changed the summation index.  
Now, assuming $1 \le m$, we can use the fundamental reursive identity of Bernoulli numbers
(the "standard" definition $B_{\,j} ^ -$)
$$
\sum\limits_{0 \le \,j\, \le \,n - m} {\left( \matrix{
  n + 1 - m \cr 
  j \cr}  \right)B_{\,j} }  = \delta _{\,n - m,\,0} 
$$
to get
$$
 S(n,m) = \sum\limits_{1 \le \,k\, \le \,n} {{1 \over k}\binom{n}{k-1} \binom{k}{m}
 B_{\,k - m}^{\, - } }  = \delta _{\,n,\;m} \quad \left| {\;1 \le n,m} \right.
$$
