An identity involving binomial coefficients and Stirling numbers of both kinds I calculated, using Mathematica, that for $4\leq k \leq 100$,
$$ \sum_{j=k}^{2k} \sum_{i=j+1-k}^j (-1)^j 2^{j-i} \binom{2k}{j} S(j,i) s(i,j+1-k) = 0,$$
where $s(i,j)$ and $S(i,j)$ are Stirling numbers of the first and second kinds, respectively.
Here the code:

F[k_] := Sum[(-1)^j 2^(j - i) Binomial[2 k, j] StirlingS2[j, 
       i] (StirlingS1[i, j + 1 - k]), {j, k, 2 k}, {i, j - k + 1, j}];
Table[F[k], {k, 4, 100}]

How do I prove it holds for all $k \geq 4$ ?
 A: Just a first step (too long for a comment).
Let me change your symbology and put
$$
S(n) = \sum\limits_{j = n}^{2n} {\sum\limits_{k = j + 1 - n}^j {\left( { - 1} \right)^{\,j} 2^{\,j - k} 
  \left( \matrix{2n \cr  j \cr}  \right)
 \left\{ \matrix{  j \cr   k \cr}  \right\}
\left[ \matrix{  k \cr j + 1 - n \cr}  \right]} }  
$$
where the brackets indicate  respectively the binomial, Stirling 2nd kind, un-signed Stirling 1st kind.
Note that if you extend the lower limit to start from $0$ then you get a cleaner result:
$S(n)=0$ for any $0 \le n$.
And since the sum bounds are implicit in the Binomial and Stirling numbers
we can plainly omit them, thus simplifying the algebraic operations.
$$
\eqalign{
  & S(n) = \sum\limits_{0\,\, \le \,j\, \le \,2n} {\;\sum\limits_{0\,\, \le \,k\, \le \,j} {
\left( { - 1} \right)^{\,j} 2^{\,j - k}
 \left( \matrix{  2n \cr   j \cr}  \right)\left\{ \matrix{  j \cr   k \cr}  \right\}\left[ \matrix{  k \cr   j + 1 - n \cr}  \right]}
 }  =   \cr 
  &  = \sum\limits_{\left( {0\,\, \le } \right)\,j\, \le \,\left( {2n} \right)} {\;\sum\limits_{\left( {0\,\, \le } \right)\,k\,\left( { \le \,j} \right)} {
\left( { - 1} \right)^{\,j} 2^{\,j - k}
 \left( \matrix{  2n \cr   j \cr}  \right)\left\{ \matrix{  j \cr   k \cr}  \right\}\left[ \matrix{  k \cr   j + 1 - n \cr}  \right]}
 } 
 = 0\quad \left| {\,0 \le n} \right. \cr
} 
$$
