Five. (Full marks: 15 points) From the 91 natural numbers $1,2,3, \cdots, 90,91$, select $k$ numbers such that there must be two natural numbers $p, q$ satisfying $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$. Determine the minimum value of the natural number $k$, and explain your reasoning.
Five, divide the 91 natural numbers from 1 to 91 into nine groups, such that the ratio of any two natural numbers in each group is no less than $\frac{2}{3}$ and no more than $\frac{3}{2}$, and the division is as follows:
$$
\begin{array}{l}
A_{1}=\{1\}, A_{2}=\{2,3\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{7,8,9,10\}, \\
A_{5}=\{11,12,13,14,15,16\}, \\
A_{6}=\{17,18,19,20,21,22,23,24,25\}, \\
A_{7}=\{26,27,28,29, \cdots, 39\}, \\
A_{8}=\{40,41,42,43, \cdots, 58,59,60\}, \\
A_{9}=\{61,62,63, \cdots, 89,90,91\} .
\end{array}
$$

If any nine numbers are taken from these 91 numbers, for example, one number from each of $A_{1}, A_{2}, \cdots, A_{9}$, such as $1,3,6,10,16,25,39,60,91$, the ratio of any two of these nine numbers is less than $\frac{2}{3}$ or greater than $\frac{3}{2}$. Therefore, $k > 9$. So $k \leqslant 10$. When $k=10$, if any 10 numbers are taken from the 91 natural numbers from 1 to 91, according to the pigeonhole principle, there must be two numbers in the same $A_{i}$. For example, these two numbers are $p$ and $q$, and $p<q$, then $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$ holds.
Therefore, the minimum value of $k$ is 10.