jmc

If $x^{101}+Ax+B$ is divisible by $x^2+x+1,$ then $x^{101}+Ax+B$ must be equal to 0 any time $x$ is a root of $x^2+x+1=0.$

Let $\omega$ be a root of $x^2+x+1=0,$ so ${\omega }^2+\omega +1=0.$ Then $$(\omega -1)({\omega }^2+\omega +1)=0,$$or ${\omega }^3-1=0,$ which means ${\omega }^3=1.$

By the Factor Theorem, $${\omega }^{101}+A\omega +B=0.$$We have that ${\omega }^{101}={\omega }^{3\cdot 33+2}=({\omega }^3{)}^{33}\cdot {\omega }^2={\omega }^2,$ so $$\begin{array}{rl}{\omega }^{101}+A\omega +B& ={\omega }^2+A\omega +B\\ & =(-\omega -1)+A\omega +B\\ & =(A-1)\omega +(B-1)\\ & =0.\end{array}$$Since $\omega$ is a nonreal complex number, we must have $A=1$ and $B=1,$ so $A+B=\boxed{2}.$