Let $$ A= {1\over 1\cdot 2}+\color{red}{1\over 3\cdot 4}+...+ {1\over 1997\cdot 1998}$$
and $$B= {1\over 1000\cdot 1998}+{1\over 1001\cdot 1997}+...+ {1\over 1997\cdot 1001}+{1\over 1998\cdot 1000}$$
Prove that $A\over B$ is an integer.
I could only find that $$ A= 1-{1\over 1998}$$ using standard trick ${1\over x(x+1)} = {1\over x}-{1\over x+1}$. But I could not find answer for the second one. It is supposed to be a task for 15 years old (Romanian) children!
Edit I write it down wrong, so the $A$ is not correctly calculated. And it is different $A$ as in suggested duplicate.
When denote $$ H(n) = \sum_{k=1}^n \dfrac{1}{k}, $$ then $A$ could be expressed as $$ A = \left(\dfrac{1}{1}-\dfrac{1}{2}\right) + \left(\dfrac{1}{3}-\dfrac{1}{4}\right) + \ldots + \left(\dfrac{1}{1997}-\dfrac{1}{1998}\right) \\ = H(1998) - 2\left(\dfrac{1}{2}+\dfrac{1}{4}+\ldots+\dfrac{1}{1998}\right)\\ = H(1998) - H(999). $$
Based on lab bhattacharjee's comment, $B$ could be expressed as $$ B = \sum_{r=1000}^{1998}\dfrac1{r(2998-r)}\\=\dfrac1{2998}\sum_{r=1000}^{1998}\left(\dfrac1r+\dfrac1{2998-r}\right)\\=\dfrac2{2998}\sum_{r=1000}^{1998}\dfrac1r\\ = \dfrac{1}{1499} \Bigl(H(1998)-H(999)\Bigr). $$
Therefore $\dfrac{A}{B}=1499$.