jmc

Noticing that $190\equiv 0(\mathrm{mod}19)$ and $-200+190=-10$, we can say that $$-200\equiv n(\mathrm{mod}19)$$if and only if $$-10\equiv n(\mathrm{mod}19).$$This is not in the range $0\le n<19$, but adding 19 again gives $$9\equiv n(\mathrm{mod}19).$$The answer is $n=\boxed{9}$.