An algebra problem by Julian Yu

$\dfrac{1+2+3+...+n}{1^2+2^2+3^2+...+n^2}=\dfrac{\frac{n(n+1)}{2}}{\frac{n(n+1)(2n+1)}{6}}=\dfrac{3}{2n+1}>\dfrac{1}{2017}$ .

Therefore, $6051>2n+1\implies n<3025$ . Therefore, the largest integer $n$ is $3024$ .