Find the number of positive integers $$n$$ less than or equal to $$100$$, which are divisible by $$3$$ but are not divisible by $$2$$.

Multiples of $$3$$ that lie in $$[1,100]$$ are
$$3,6,9,\dots ,99$$.
The count of these numbers is given by $$\left\lfloor \frac{100}{3} \right\rfloor = 33$$.

A number divisible by both $$3$$ and $$2$$ must be divisible by their LCM, $$6$$. Multiples of $$6$$ in $$[1,100]$$ are
$$6,12,18,\dots ,96$$.
The count of these numbers is $$\left\lfloor \frac{100}{6} \right\rfloor = 16$$.

Numbers that are divisible by $$3$$ but not by $$2$$ are obtained by subtracting the above counts:
$$33 - 16 = 17$$.

Therefore, the required number of positive integers $$n \le 100$$ that satisfy the given condition equals $$17$$.

Answer: 17