jmc

We can re-write the requirements of this problem as: $$\begin{array}{rl}k& \equiv 1(\mathrm{mod}3)\\ k& \equiv 1(\mathrm{mod}4),\\ k& \equiv 1(\mathrm{mod}5),\\ k& \equiv 1(\mathrm{mod}6),\\ k& \equiv 1(\mathrm{mod}7).\end{array}$$In other words, $k-1$ is divisible by 3, 4, 5, 6, and 7. We have $$\begin{array}{rl}\text{lcm}[3,4,5,6,7]& =\text{lcm}[3,4,5,7]\\ & =3\cdot 4\cdot 5\cdot 7\\ & =420,\end{array}$$and thus $k-1$ must be divisible by $420$. The only two multiples of $420$ under $1000$ are $420$ and $840$, so the largest possible value of $k$ is $840+1=\boxed{841}$.