Cosecant Trig Sum

Note that $\begin{aligned} \sin(2n+1)x & = \; \mathfrak{Im}\left[(\cos x + i \sin x)^{2n+1}\right] \; = \; \sum_{j=0}^n \binom{2n+1}{2j+1}(-1)^j \cos^{2n-2j}x \sin^{2j+1}x \\ & = \; \sin^{2n+1}x F_n(\cot^2x) \end{aligned}$ where $F_n(X) \; = \; \sum_{j=0}^n \binom{2n+1}{2j+1}(-1)^j X^{n-j} \; = \; \binom{2n+1}{1}X^n - \binom{2n+1}{3}X^{n-1} + \cdots$ We see that $F_n(\cot^2x) = 0$ whenever $\sin(2n+1)x=0$ but $\sin x \neq 0$ . Thus the roots of $F_n(X)$ are $\cot^2\tfrac{k\pi}{2n+1} \hspace{2cm} 1 \le k \le n$ Thus $\begin{aligned} \sum_{k=1}^{2n}\csc^2\tfrac{k\pi}{2n+1} & = \; 2\sum_{k=1}^n\csc^2\tfrac{k\pi}{2n+1} \; = \; 2\left(n + \sum_{k=1}^n \cot^2\tfrac{k\pi}{2n+1}\right) \\ & = \; 2\left(n + \binom{2n+1}{3} \times \binom{2n+1}{1}^{-1}\right) \; = \; 2\left[n + \tfrac13n(2n-1)\right] \\ & = \; \tfrac43n(n+1) \end{aligned}$ With $n=2019$ , the answer is $\boxed{5437840}$ .