$2 \cdot 6$ In the set $M$ of the first 100 odd numbers $1,3, \cdots, 199$, select a subset such that no number in the subset can divide another. How many elements can this subset have at most?
[Solution] Let
$$
S=\{2 n+1 \mid 33 \leqslant n \leqslant 99\},
$$

then $S$ has 67 elements and satisfies the requirements of the problem. If there exist $a, b \in S$ such that $a \mid b$, then since $a, b$ are both odd, the quotient $\frac{b}{a}$ is also odd. Thus $\frac{b}{a} \geqslant 3, b \geqslant 3 a \geqslant 201$, which is a contradiction.
On the other hand, let
$$
T_{m}=\left\{k \mid k=3^{r} m,(m, 3)=1, k \in M\right\},
$$

then $1 \leqslant m \leqslant 199,(m, 3)=1,(m, 2)=1$. There are 67 such $m$. Let $A$ be any subset of $M$ with more than 67 elements. By the pigeonhole principle, there must be two elements belonging to the same $T_{m_{0}}$, thus the larger one must be divisible by the smaller one. Therefore, any subset with more than 67 elements does not satisfy the requirements of the problem. Hence, the maximum number of elements is 67.