China Southeastern Mathematical Olympiad

${\sum }_{k=1}^{l}k=S_l$, ${\sum }_{k=l+1}^{m}k=S_m-S_l$, ${\sum }_{k=m+1}^{n}k=S_n-S_m$, form a geometric sequence in order. Then $$S_l(S_n-S_m)=(S_m-S_l{)}^2,\stackrel{◯}{1}$$ that is $S_l(S_n+S_m-S_l)=S_m^2$. Thus, $S_l\mid S_m^2$, that is $$2l(l+1)\mid m^2(m+1{)}^2.$$ Let $m+1=l(l+1)$ and take $l=3$. Then $m=11$ and $S_l=S_3=6$, $S_m=S_{11}=66$. Substitute it into ①, we have $S_n=666$, that is $\frac{n(n+1)}{2}=666$. So $n=36$. Therefore, $(l,m,n)=(3,11,36)$ is a solution satisfying the condition. □