smc

According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x)=Q(x)(x-19)+99=R(x)(x-99)+19$. From the last equality we get $Q(x)(x-19)+80=R(x)(x-99)$. The value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on the left hand side as well. Substituting, we get $Q(99)(99-19)+80=0$, from which $Q(99)=-1$. This means that $99$ is a root of the polynomial $Q(x)+1$. In other words, there is a polynomial $S(x)$ such that $Q(x)+1=S(x)(x-99)$. Substituting this into the original formula for $P(x)$ we get $$P(x)=Q(x)(x-19)+99=(S(x)(x-99)-1)(x-19)+99=$$ $$=S(x)(x-99)(x-19)-(x-19)+99$$ Therefore when $P(x)$ is divided by $(x-19)(x-99)$, the remainder is $\boxed{-x+118}$.