we just have to check n=1,2,3 from Cauchy schwartz inequality, $(a_1+2a_2+......na_n)(\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+ ... +\frac{n}{a_n})\ge (1+2+3+....n)^2$ solving this we get $6n\times (2+1/n)\ge(\frac{n(n+1)}{2})^2$ and we get $n^4+2n^3+n^2-48n-24\le 0$ we can check that when n>3 then this inequality cant be satisfied We can easily check that n=1 doesnt work $a_1=6$ and this doesnt work when n=2 we can prove this doesnt work as $x+2y=12$ and $\frac{1}{x}+\frac{2}{y}=\frac{5}{2}$ and solving this quadratic gives us a contradiction n=3 works example: $a_1=6$ $a_2=3$ $a_3=2$ and so n=3 is the only solution