9. Let sets $A$ and $B$ satisfy $A \cup B=\{1,2, \ldots, 10\}$, if $A \cap B=\varnothing$, if the number of elements in set $A$ is not an element of set $A$, and the number of elements in set $B$ is not an element of set $B$, then the number of all sets $A$ that satisfy the conditions is $\qquad$
Let $|A|=k$, then $|B|=10-k, k \neq 5$, otherwise $5 \in A \cap B$,
thus according to the problem: $k \in B, 10-k \in A$, at this point, the remaining $k-1$ elements in $A$ have $\mathrm{C}_{8}^{k-1}$ choices, and the remaining $9-k$ elements must belong to set $B$.

Therefore, the number of sets $A$ that satisfy the condition is $m=\sum_{k=1}^{9} \mathrm{C}_{8}^{k-1}-\mathrm{C}_{8}^{5-1}=2^{8}-70=256-70=186$.
