Cho n là số nguyên dương, tìm n sao cho:

\[{{1}^{2}}{{\log }_{a}}2019+{{2}^{2}}{{\log }_{\sqrt{a}}}2019+...+{{n}^{2}}{{\log }_{\sqrt[n]{a}}}2019={{1010}^{2}}\times {{2019}^{2}}{{\log }_{a}}2019\]