imc

Since the least common multiple $\mathrm{lcm}(4,6)=12$, the elements that are common to $S$ and $T$ must be multiples of $12$. Since $4\cdot 2005=8020$ and $6\cdot 2005=12030$, several multiples of $12$ that are in $T$ won't be in $S$, but all multiples of $12$ that are in $S$ will be in $T$. So we just need to find the number of multiples of $12$ that are in $S$. Since $4\cdot 3=12$, every $3$rd element of $S$ will be a multiple of $12$. Therefore the answer is $\lfloor \frac{2005}{3}\rfloor =\boxed{668}$