【Question 11】8 different non-zero natural numbers are arranged in ascending order. The average of the first 3 numbers is 9, the average of the 8 numbers is 19, and the average of the last 3 numbers is 29. What is the maximum difference between the second largest and the second smallest number? $\qquad$ _.

Analysis:
Let the 8 distinct non-zero natural numbers be $a, b, c, d, e, f, g, h$,

and $0<a<b<c<d<e<f<g<h$

From the problem, we know $a+b+c=27, f+g+h=87$, so $d+e=19 \times 8-27-87=38$

To ensure the difference between $g$ and $b$ is maximized, $g$ should be as large as possible, and $b$ should be as small as possible.
To make $b$ as small as possible, $c$ should be as large as possible. Since $c<d$ and $d$ can be at most 18 (because $d+e=38=18+20$), $c$ can be at most 17. Therefore, $a+b=27-c=10=4+6, a$ can be at most 4, so $b$ is at least 6.

To make $g$ as large as possible, $f$ should be as small as possible. Since $e<f$ and $e$ can be at least 20 (because $d+e=38=18+20$), $f$ can be at least 21. Therefore, $g+h=87-f=66=32+34, h$ can be at least 34, so $g$ can be at most 32.

Thus, the maximum difference between the second largest number and the second smallest number is $g-b=32-6=26$.