10. (10 points) Among the following $2 n$ numbers, what is the maximum number of numbers that can be selected such that the ratio of any two selected numbers is not 2 or $\frac{1}{2}$?
$3, 3 \times 2, 3 \times 2^{2}, 3 \times 2^{3}, 3 \times 2^{4}, 3 \times 2^{5}, \cdots, 3 \times 2^{2 n-1}$.
【Analysis】Using proof by contradiction, assume that the number of selected numbers is more than $n$, then it is possible to pick $n+1$ numbers from them, and derive a contradictory conclusion through reasoning. Solve accordingly.

【Solution】Solution: The maximum number of numbers that can be selected is $n$, such as $3,3 \times 2^{2}, \cdots 3 \times 2^{2 n-1}$. If the number of selected numbers is more than $n$, then it is possible to pick $n+1$ numbers as follows: $2 l_{1} \times 3,2 l_{2} \times 3,2 l_{3} \times 3, \cdots$, $2 l_{n+1} \times 3, \quad l_{1}l_{n+1} .-l_{1}=\left(l_{n+1} .-l_{n}\right)+\left(l_{n}-l_{n-1}\right)+\cdots+\left(l_{2}-l_{1}\right) \geqslant_{\mathrm{n}}$ numbers $=2 n$. $2 n-1>2 n$, which is impossible.

Therefore, the number of selected numbers cannot exceed $n$.