11. (This sub-question is worth 20 points) It is known that any positive integer $n$ can be uniquely represented as
$$
n=a_{0}+a_{1} 9+a_{2} 9^{2}+\cdots+a_{m} 9^{m}
$$

where $m$ is a non-negative integer, $a_{j} \in\{0,1, \cdots, 8\}(j=0,1, \cdots, m-1), a_{m} \in\{1, \cdots, 8\}$. Find the sum of all positive integers $n$ for which the sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{m}$ in $(*)$ is strictly increasing or strictly decreasing.
Let $A$ and $B$ denote the sets of all positive integers that form strictly increasing and decreasing sequences, respectively, in $(*)$. The symbol $S(M)$ represents the sum of all numbers in the set $M$, and the positive integers satisfying $(*)$ are denoted as:
$$
n=\overline{a_{m} a_{m-1} \cdots a_{1} a_{0}}
$$

Divide the set $A$ into two disjoint subsets $A_{0}=\left\{n \in A \mid a_{0}=0\right\}$ and $A_{1}=\left\{n \in A \mid a_{0} \neq 0\right\}$. We have:
$$
S(A)=S\left(A_{0}\right)+S\left(A_{1}\right)
$$
For all $n \in A_{1}$, let $f(n)=9 n \in A_{0}$, then $f$ is a bijection from $A_{1}$ to $A_{0}$, thus: $S\left(A_{0}\right)=9 S\left(A_{1}\right)$, and therefore:
$$
S(A)=10 S\left(A_{1}\right)
$$
5 points

For all $a=\overline{a_{m} a_{m-1} \cdots a_{0}} \in B$, let $b=g(a)=\overline{\left(9-a_{m}\right)\left(9-a_{m-1}\right) \cdots\left(9-a_{0}\right)} \in A_{1}$, then $g$ is a bijection from $B$ to $A_{1}$, where $a+b=9^{m+1}+9^{m}+\cdots+9=\frac{9}{8}\left(9^{m+1}-1\right)$.

Since $B=\left\{\overline{a_{m} a_{m-1} \cdots a_{0}} \mid 1 \leq a_{m}<a_{m-1}<\cdots<a_{0} \leq 8, m=0,1, \cdots, 7\right\}$, $B$ contains $\sum_{m=0}^{7} C_{8}^{m+1}$ elements, thus:
$$
S(B)+S\left(A_{1}\right)=\frac{9}{8} \sum_{m=0}^{7} C_{8}^{m+1}\left(9^{m+1}-1\right)=\frac{9}{8} \sum_{k=0}^{8} C_{8}^{k} 9^{k}-\frac{9}{8} \sum_{k=0}^{8} C_{8}^{k}=\frac{9}{8}\left(10^{8}-2^{8}\right) \quad 10 \text { points }
$$

Let $A_{2}$ denote the set of all positive integers in $A$ with the highest digit $a_{m}=8$, and let $A_{3}$ be the set of all other numbers in $A$ and zero. Then:
$$
S(A)=S\left(A_{2}\right)+S\left(A_{3}\right)
$$

For all $a=\overline{a_{m} a_{m-1} \cdots a_{0}} \in B$, let $b=\sigma(a)=\overline{\left(8-a_{m}\right)\left(8-a_{m-1}\right) \cdots\left(8-a_{0}\right)} \in A_{3}$, then $\sigma$ is a bijection from $B$ to $A_{3}$, where $a+b=8 \cdot 9^{m}+8 \cdot 9^{n-1}+\cdots+8=9^{m+1}-1$, so:
$$
S(B)+S\left(A_{3}\right)=\sum_{m=0}^{7} C_{8}^{m+1}\left(9^{m+1}-1\right)=\sum_{k=0}^{8} C_{8}^{k}\left(9^{k}-1\right)=10^{8}-2^{8}
$$

Finally, for all $a=\overline{8 a_{m} \cdots a_{0}} \in A_{2}-\{8\}$, let $b=\tau(a)=\overline{\left(8-a_{m}\right) \cdots\left(8-a_{0}\right)} \in B$, then $\tau$ is a bijection from $A_{2}-\{8\}$ to $B$, where $a+b=8 \cdot 9^{n+1}+8 \cdot 9^{m}+\cdots+8=9^{m+2}-1$, so:
$$
S(B)+S\left(A_{2}\right)=8+\sum_{m=0}^{7} C_{8}^{m+1}\left(9^{m+2}-1\right)=8+\sum_{k=1}^{8} C_{8}^{k}\left(9^{k+1}-1\right)=9 \cdot 10^{8}-2^{8}
$$

Thus, we have the system of equations:
$$
\left\{\begin{array}{l}S(B)+\frac{1}{10} S(A)=\frac{9}{8}\left(10^{8}-2^{8}\right) \\ 2 S(B)+S(A)=10^{9}-2^{9}\end{array}\right.
$$
Solving this system, we get: $S(A)=\frac{31}{32} 10^{9}+80=968750080, S(B)=15624704$.
Since $A$ and $B$ both contain $1,2, \cdots, 8$, the sum of the required positive integers is $S(A)+S(B)-36=984374748$. 20 points