jmc

Among five consecutive odd numbers, at least one is divisible by 3 and exactly one is divisible by 5, so the product is always divisible by 15. The cases $n=2$, $n=10$, and $n=12$ demonstrate that no larger common divisor is possible, since $\boxed{15}$ is the greatest common divisor of $3\cdot 5\cdot 7\cdot 9\cdot 11$, $11\cdot 13\cdot 15\cdot 17\cdot 19$, and $13\cdot 15\cdot 17\cdot 19\cdot 21$.