8. Let $x_{1}, x_{2}, x_{3}, \cdots, x_{9}$ be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots x_{9}=220$, then when $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is
(A) 8 ;
(B) 9 ;
(C) 10 ;
(D) 11 .
8. (B)
Solution (1) First, prove that $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
If not, assume $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$, then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant 27, x_{8} \geqslant 28, x_{9} \geqslant 29$. Therefore, $x_{1}+x_{2}+\cdots+x_{9}>220$. This contradicts the assumption.
(2) If we take $x_{1}=20, x_{2}=21, x_{3}=22, x_{4}=23, x_{5}=24$, then $x_{1}+x_{2}+\cdots+x_{5}=110$. Therefore, when $x_{1}+x_{2}+\cdots+x_{5}$ reaches its maximum value, the largest value of $x_{1}$ is 20.
(3) If we take $x_{6}=26, x_{7}=27, x_{8}=28, x_{9}=29$, then $x_{6}+x_{7}+x_{8}+x_{9}=110$. Therefore, when $x_{1}+x_{2}+\cdots+x_{5}$ reaches its maximum value, the smallest value of $x_{9}$ is 29. Thus, the minimum value of $x_{9}-x_{1}$ is $29-20=9$.