1. B divides the 99 positive integers from 1 to 99 into 6 groups, such that the ratio of any two numbers in each group is within the closed interval $\left[\frac{1}{2}, 2\right]$, and the number of elements in each group is as large as possible. The groups are as follows: $A_{1}=\{1,2\}, A_{2}=\{3,4,5,6\}, A_{3}=\{7,8, \cdots, 14\}, A_{4}=\{15$, $16, \cdots, 30\}, A_{5}=\{31,32, \cdots, 62\}, A_{6}=\{63,64, \cdots, 99\}$. For example, if we take one number from each of $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$, such as $1,3,7,15,31,63$, the ratio of any 2 of these 6 numbers is not within the closed interval $\left[\frac{1}{2}, 2\right]$, indicating that $k>$ 6. Therefore, $k \geqslant 7$. When $k=7$, if we take any 7 numbers from the 99 numbers, by the pigeonhole principle, there must be 2 numbers belonging to the same $A_{i}(1 \leqslant i \leqslant$ 6), and the ratio of these 2 numbers is within the closed interval $\left[\frac{1}{2}, 2\right]$. Therefore, the minimum value of $k$ is 7.
1. In the 99 positive integers $1,2, \cdots, 99$, if any $k$ numbers are taken out, such that there must be two numbers $a, b(a \neq b)$ satisfying $\frac{1}{2} \leqslant \frac{b}{a} \leqslant 2$. Then the smallest possible value of $k$ is
A. 6
B. 7
C. 8
D. 9