13.6 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 the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is
(A) 8.
(B) 9.
(C) 10.
(D) 11.
(China Junior High School Mathematics League, 1992)
[Solution 1] (1) First, prove $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$. Use proof by contradiction.
Assume $x_{1}+x_{2}+\cdots+x_{5}>110$, then $x_{6}+x_{7}+x_{8}+x_{9}<110$,
and $x_{7} \geqslant x_{6}+1, \quad x_{8} \geqslant x_{7}+1, \quad x_{9} \geqslant x_{8}+1$,
thus $\quad x_{6}+\left(x_{6}+1\right)+\left(x_{6}+2\right)+\left(x_{6}+3\right)$
$$
\leqslant x_{6}+x_{7}+x_{8}+x_{9}<110 \text {. }
$$

Then $x_{6}<\frac{110-6}{4}=26$, and since $x_{6}$ is a positive integer, thus $x_{6} \leqslant 25$.
This implies $x_{5} \leqslant 24, \quad x_{4} \leqslant 23, x_{3} \leqslant 22, x_{2} \leqslant 21, \quad x_{1} \leqslant 20$.
And $20+21+22+23+24=110$.
Therefore, $x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \leqslant 110$, which contradicts the initial assumption!
(2) If we take $x_{1}=20, x_{2}=21, x_{3}=22, x_{4}=23, x_{5}=24$,
then $x_{1}+x_{2}+\cdots+x_{5}=110$.
Hence, when $x_{1}+x_{2}+\cdots+x_{5}$ reaches its maximum value, the maximum value of $x_{1}$ is 20.
(3) When $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=110$, if we take $x_{6}=26, x_{7}=27$, $x_{8}=28, x_{9}=29$, then $x_{6}+x_{7}+x_{8}+x_{9}=110$, satisfying the problem's conditions.
Thus, when $x_{1}+x_{2}+\cdots+x_{5}$ reaches its maximum value, the minimum value of $x_{9}$ is 29.
Therefore, when $x_{1}+x_{2}+\cdots+x_{5}$ reaches its maximum value, the minimum value of $x_{9}-x_{1}$ is 9.
Hence, the answer is (B).
[Solution 2] By $x_{9} \geqslant x_{8}+1 \geqslant x_{7}+1 \geqslant \cdots \geqslant x_{1}+8$, then $x_{9}-x_{1} \geqslant 8$.
If $x_{9}-x_{1}=8$, then $x_{1}, x_{2}, \cdots, x_{9}$ form an arithmetic sequence with a common difference of 1, and their sum is 220, i.e., $9 x_{5}=220, x_{5}$ is not an integer, which is a contradiction! Hence, $x_{9}-x_{1} \neq 8$.
Thus, $x_{9}-x_{1} \geqslant 9$.
If we take $x_{1}=20, \quad x_{2}=21, \quad x_{3}=22, \quad x_{4}=23, \quad x_{5}=24, \quad x_{6}=$
$26, x_{7}=27, x_{8}=28, x_{9}=29$,
then $\sum_{i=1}^{9} x_{i}=220$, and $x_{9}-x_{1}=9$.
Thus, the minimum value of $x_{9}-x_{1}$ is 9.
Hence, the answer is (B).