8. For some subsets of the set $\{1,2,3, \cdots, 100\}$, no number is twice another number. The maximum number of elements in such a subset is $\qquad$ .
8. 67 .
$$
\text{Let } A_{1}=\{51,52, \cdots, 100\}, A_{2}=\{26,27, \cdots, 50\}, A_{3}=\{13,14, \cdots, 25\}, A_{4}=\{7, 
$$
$8,9,10,11,12\}, A_{5}=\{4,5,6\}, A_{6}=\{2,3\}, A_{7}=\{1\}$, then $A_{1} \cup A_{3} \cup A_{5} \cup A_{7}$ has $50+$ $13+3+1=67$ (elements), none of which is twice another. If $B \subseteq\{1,2, \cdots, 100\}$, where each number is not twice another, then when $a \in B \cap A_{2}$, $2 a \notin B$. Therefore, $\left|B \cap A_{4}\right|+\left|B \cap A_{3}\right| \leqslant 13$, $\left|B \cap A_{6}\right|+\left|B \cap A_{5}\right| \leqslant 3$. Therefore, $|B| \leqslant 67$, hence the maximum is 67.