jmc

Let $a=a_1,$ and let $d$ be the common difference, so $$S_n=\frac{2a+(n-1)d}{2}\cdot n.$$Then $$\begin{array}{rl}{T}_n& =\sum _{k=1}^n\left(\frac{2a+(k-1)d}{2}\cdot k\right)\\ & =\sum _{k=1}^n\left(\left(a-\frac{d}{2}\right)k+\frac{d}{2}{k}^2\right)\\ & =\left(a-\frac{d}{2}\right)\sum _{k=1}^{n}k+\frac{d}{2}\sum _{k=1}^n{k}^2\\ & =\left(a-\frac{d}{2}\right)\cdot \frac{n(n+1)}{2}+\frac{d}{2}\cdot \frac{n(n+1)(2n+1)}{6}\\ & =\frac{n(n+1)(3a+(n-1)d)}{6}.\end{array}$$We are told the value of $$S_{2019}=\frac{2a+2018d}{2}\cdot 2019=2019(a+1009d),$$which means the value of $a+1009d$ is uniquely determined. Then the value of $3(a+1009d)=3a+3027d$ is uniquely determined. Thus, we can determine $T_n$ for $n=3027+1=\boxed{3028}.$