By definition (at least the one I'm used to), the Stirling numbers of the first kind $S_1(n,k)$ satisfy
$$ (x)_n = \sum_{k=0}^n S_1(n,k) x^k, $$ and the Stirling numbers of the second kind $S_2(n,k)$ satisfy
$$ x^n = \sum_{k=0}^n S_2(n,k) (x)_k $$ where $(x)_n=x(x-1)\cdots(x-n+1)$ is the falling factorial (with $(x)_0=1$). Combining these yields
$$ x^n = \sum_{k=0}^n \sum_{l=0}^k S_2(n,k) S_1(k,l) x^l $$
$$ = \sum_{l=0}^n x^l \left( \sum_{k=l}^n S_2(n,k) S_1(k,l) \right). $$
Comparing powers of $x$, we see that
$$ \sum_{k=l}^n S_2(n,k) S_1(k,l) = \left\{
\begin{array}{lr}
1 & \mathrm{ if}\;\; l=n \\
0 & \mathrm{ if}\;\; l\neq n
\end{array}
\right.$$
$$ = \delta_{ln}.$$
If we define $S_{\nu}(a,b)=0$ for $a<b$, then this is just the product of the $n^{th}$ row of $S_N$ and the $l^{th}$ column of $s_N$, i.e. the $(n,l)^{th}$ element of the matrix product $S_N s_N$ is $\delta_{nl}$. Thus their product is the identity matrix, and hence they're matrix inverses of eachother.
[Edit] Another way (less computation, slightly more hand-wavy) to see this is that $S_N$ is the change of basis matrix (for the space of polynomials) from $\{1,x,x^2,\dots\}$ to $\{(x)_0, (x)_1, \dots\}$, and $s_N$ is the change of basis matrix going the other way. Hence the linear transformation $S_Ns_N$ takes the coefficients in terms of $\{x^i\}$ to coefficients in terms of $\{(x)_i\}$ and then back to $\{x^i\}$, i.e. it's the identity.
"Concrete Mathematics (what else?) - Eulerian Numbers" - says:
"Second-order Eulerian numbers are important chiefly because of their connection with Stirling numbers"
Eq. (6.43) therein gives
$$
\left\{ \matrix{ x \cr x - n \cr} \right\}
= \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{
n \cr
k \cr} \right\rangle } \right\rangle } \binom{x+n-1-k}{2n}
\quad \left| \matrix{
\;0 \le n \in Z \hfill \cr
\;x \in C \hfill \cr} \right.
$$
which easily reduce to yours, and can be extended to define Stirling No. of 2nd kind, of the indicated form,
to complex values of $x$.
Interestingly, also given is a twin one for the Stirling No. of 1st kind
$$
\left[ \matrix{ x \cr x - n \cr} \right]
= \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{
n \cr k \cr} \right\rangle } \right\rangle } \binom{x+k}{2n}
\quad \left| \matrix{
\;0 \le n \in Z \hfill \cr
\;x \in C \hfill \cr} \right.
$$
Best Answer
Since $s + 1 \leq i \leq n$, we see that $i>s \, \, \, \forall s < n$. Hence $\binom{s}{i}=0$. But each term of your sum is the product of $\binom{s}{i}$ and other values, which means each term will be zero. So the entire sum will be zero.