jmc

Since any positive integer(expressed in base ten) is some multiple of $5$ plus its last digit, its remainder when divided by $5$ can be obtained by knowing its last digit.

Note that $1999^1$ ends in $9,$ $1999^2$ ends in $1,$ $1999^3$ ends in $9,$ $1999^4$ ends in $1,$ and this alternation of $9$ and $1$ endings continues with all even powers ending in $1.$ Therefore, the remainder when $1999^{2000}$ is divided by $5$ is $\boxed{1}.$