The answer is $6$. For $n=5$, just consider $45\nmid17\times19\times21\times23\times25$. Since for any consecutive $6$ odd positive integers, at least $2(=\lfloor\frac63\rfloor)$ of which are divisible by $3$, and at least $1=(=\lfloor\frac65\rfloor)$ of which is divisible by $5$, therefore the product of which is divisible by $45$.

the mininum number of $n$ is $6$. It is obvious that $n>4$ because if it is not, there could not be a number that divdes $5$ in the integers. It is impossible when $n=5$ ($ 35,37,39,41,43$ for example) it is always possible when $n=6$ because there are always two numbers that divede $3$. So, $n=6$.