The number of nonempty subsets $S\subseteq \{-11,-10,-9,-8,\mathrm{...........8},9,10,11\}$ that satisfy $\left|S\right|+\min \left(S\right).\max \left(S\right)=0$ is R, then R is divisible by
($\left|S\right|$ denotes number of elements of set S, min (S) denotes minimum element of set S and max(S) denotes maximum element of set S)

Since min(S). max(S) < 0, we must have min(S) = –a and max(S) = b for some positive integers a and b. Given a and b, there are $\left|S\right|-2=ab-2$ elements left to choose, which must come from the set  $\{-a+1,-a+\mathrm{2.........},b-2,b-1\}$, which has size a+b–1. Therefore, the number of possibilities for a given a, b are $\left(\begin{array}{c}a+b-1\\ ab-2\end{array}\right)$.
In particular, we must have $ab-2\le a+b-1\iff (a-1)(b-1)\le 2$. The possibilities for (a, b) to (1, n) and (n, 1) for positive integers $2\le n\le 11$ (n = 1 case is impossible), and three extra possibilities: (2,2), (2,3), and 3, 2). In the first case, the number of possible sets is
$2(\left(\begin{array}{c}2\\ 0\end{array}\right)+\left(\begin{array}{c}3\\ 1\end{array}\right)+\mathrm{......}+\left(\begin{array}{c}10\\ 8\end{array}\right)+\left(\begin{array}{l}11\\ 9\end{array}\right))=2(\left(\begin{array}{c}2\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ 2\end{array}\right)+\mathrm{.....}+\left(\begin{array}{c}11\\ 2\end{array}\right))=2\left(\begin{array}{c}12\\ 3\end{array}\right)=440$ 
In the second case, the number of possible sets is
$\left(\begin{array}{c}3\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ 4\end{array}\right)+\left(\begin{array}{c}4\\ 4\end{array}\right)=5$
Thus, there are 445 sets in total.