jmc

For $n=1,$ we get $a_1=1.$ Otherwise, $$\sum _{k=1}^n{k}^2{a}_k=n^2.$$Also, $$\sum _{k=1}^{n-1}{k}^2{a}_k=(n-1{)}^2.$$Subtracting these equations, we get $$n^2{a}_n=n^2-(n-1{)}^2=2n-1,$$so $a_n=\frac{2n-1}{n^2}=\frac{2}{n}-\frac{1}{n^2}.$ Note that $a_n=1-\frac{n^2-2n+1}{n^2}=1-{\left(\frac{n-1}{n}\right)}^2$ is a decreasing function of $n.$

Also, $$a_{4035}-\frac{1}{2018}=\frac{2}{4035}-\frac{1}{4035^2}-\frac{1}{2018}=\frac{1}{4035\cdot 2018}-\frac{1}{4035^2}>0,$$and $$a_{4036}<\frac{2}{4036}=\frac{1}{2018}.$$Thus, the smallest such $n$ is $\boxed{4036}.$