jmc

Let $n=15r$. Clearly, $r>14$, because $15!$ contains 15 as a factor and all integers less than 15 as factors. If $r=15$, then $n=225$, However, $15!=15\cdot 5\cdot 3s$, so $r>15$. If $r=16$, then $n=240$. However, $15!=15\cdot 8\cdot 2t$, so $r>16$. If $r=17$, then $n=255$. Note that $f(255)=17$ because the smallest integer $k$ such that $k!$ is divisible by 17 is $k=17$, because 17 is prime. Therefore, the smallest multiple of 15 that fits the desired condition is $\boxed{n=255}$.