25~26高三下·深圳一模·第14题

$23$

$S = {82,83,84,85,86,87,89}$，$S$有$7$个元素，设$a_1 \le a_2 \le \dots \le a_8$.

$ C_8^7 = 8 > 7 $，∴$a_{i_1}+a_{i_2}+\dots+a_{i_7}$ 应有重复，设其中重复的元素为$X$.

设$a_1 + a_2 + \dots + a_8 = m$， $\therefore 7m = 82 + 83 + 84 + 85 + 86 + 87 + 89 + X$.

$\therefore 7m = 596+X$，$m$能被$596+X$整除，$X = 83$，$m = 97$.

$\therefore
\begin{cases}
82 = m – a_8 \
89 = m – a_1
\end{cases}
\Rightarrow a_1 + a_8 = 2m – 171 = 194 – 171 = 23$.

故答案为：$23$.