jmc

Note that$$\frac{\left((3!)!\right)!}{3!}=\frac{(6!)!}{6}=\frac{720!}{6}=\frac{720\cdot 719!}{6}=120\cdot 719!.$$ Because $120\cdot 719!<720!$, conclude that $n$ must be less than 720, so the maximum value of $n$ is 719. The requested value of $k+n$ is therefore $120+719=\boxed{839}$.