jmc

Let us investigate the first few powers of 19: $$\begin{array}{rl}19^1& \equiv 19(\mathrm{mod}25)\\ 19^2& \equiv 11(\mathrm{mod}25)\\ 19^3& \equiv 9(\mathrm{mod}25)\\ 19^4& \equiv 21(\mathrm{mod}25)\\ 19^5& \equiv 24(\mathrm{mod}25).\end{array}$$ At this point, we see that $19^5\equiv 24\equiv -1(\mathrm{mod}25),$ hence $19^{10}\equiv 1(\mathrm{mod}25).$ That means that $19^{1999}=19^9\cdot (19^{10}{)}^{199}\equiv 19^9(\mathrm{mod}25).$

Since $19^4\equiv 21\equiv -4(\mathrm{mod}25)$ and $19^5\equiv -1(\mathrm{mod}25),$ then $19^{1999}\equiv 19^9\equiv 4(\mathrm{mod}25),$ hence our desired remainder is $\boxed{4}.$