The computations could be continued with a closed form of Chebyshev polynomials.

An (essentially equivalent) "algebraic" alternative is as follows. \begin{align*} \sum_{k=1}^{2024}\frac{2023-2022\cos\frac{(2k-1)\pi}{2024}}{2021-2020\cos\frac{(2k-1)\pi}{2024}}&=2024\left(1+\frac1{1010}\right)-\frac1{2\cdot 1010^2}f_{2024}\left(\frac{2021}{2020}\right),\\ f_n(z)&:=\sum_{k=1}^n\frac1{z-\cos[(2k-1)\pi/n]}=g_n\left(z\pm\sqrt{z^2-1}\right),\\ g_n(z)&:=\sum_{k=1}^n\frac{2z}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{2nz}{z^2-1}\frac{z^n-1}{z^n+1}, \end{align*} and the last equality is obtained from the factorization $$\color{gray}{\left[\prod_{k=1}^n(z-e^{\pm 2k\pi i/n})=z^n-1\implies\right]}\prod_{k=1}^n(z-e^{\pm(2k-1)\pi i/n})=z^n+1.$$

After $(z-e^{it})(z-e^{-it})=z^2-2z\cos t+1$, the product is $$\prod_{k=1}^n\left(z^2-2z\cos[(2k-1)\pi/n]+1\right)=(z^n+1)^2.$$ Taking the logarithmic derivative (w.r.t. $z$), we obtain $$\sum_{k=1}^n\frac{z-\cos[(2k-1)\pi/n]}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{nz^{n-1}}{z^n+1}.$$

To get the claimed equality, multiply by $2z$, subtract $n$, and multiply by $\frac{2z}{z^2-1}$.

The computations could be continued with a closed form of Chebyshev polynomials.

An (essentially equivalent) "algebraic" alternative is as follows. $$\bbox[5pt,border:2pt solid black]{\begin{aligned} \sum_{k=1}^{2024}\frac{2023-2022\cos\frac{(2k-1)\pi}{2024}}{2021-2020\cos\frac{(2k-1)\pi}{2024}}&=2024\left(1+\frac1{1010}\right)-\frac1{2\cdot 1010^2}f_{2024}\left(\frac{2021}{2020}\right),\\ f_n(z)&:=\sum_{k=1}^n\frac1{z-\cos[(2k-1)\pi/n]}=g_n\left(z\pm\sqrt{z^2-1}\right),\\ g_n(z)&:=\sum_{k=1}^n\frac{2z}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{2nz}{z^2-1}\frac{z^n-1}{z^n+1}, \end{aligned}}$$ and the last equality is obtained from the factorization $$\color{gray}{\left[\prod_{k=1}^n(z-e^{\pm 2k\pi i/n})=z^n-1\implies\right]}\prod_{k=1}^n(z-e^{\pm(2k-1)\pi i/n})=z^n+1.$$

After $(z-e^{it})(z-e^{-it})=z^2-2z\cos t+1$, the product is $$\prod_{k=1}^n\left(z^2-2z\cos[(2k-1)\pi/n]+1\right)=(z^n+1)^2.$$ Taking the logarithmic derivative (w.r.t. $z$), we obtain $$\sum_{k=1}^n\frac{z-\cos[(2k-1)\pi/n]}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{nz^{n-1}}{z^n+1}.$$

To get the claimed equality, multiply by $2z$, subtract $n$, and multiply by $\frac{2z}{z^2-1}$.

The computations could be continued with a closed form of Chebyshev polynomials.

An (essentially equivalent) "algebraic" alternative is as follows. \begin{align*} \sum_{k=1}^{2024}\frac{2023-2022\cos\frac{(2k-1)\pi}{2024}}{2021-2020\cos\frac{(2k-1)\pi}{2024}}&=2024\left(1+\frac1{1010}\right)-\frac1{2\cdot 1010^2}f_{2024}\left(\frac{2021}{2020}\right),\\ f_n(z)&:=\sum_{k=1}^n\frac1{z-\cos[(2k-1)\pi/n]}=g_n\left(z\pm\sqrt{z^2-1}\right),\\ g_n(z)&:=\sum_{k=1}^n\frac{2z}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{2nz^n}{z^2-1}\frac{z^n-1}{z^n+1}, \end{align*} and the last equality is obtained from the factorization $$\color{gray}{\left[\prod_{k=1}^n(z-e^{\pm 2k\pi i/n})=z^n-1\implies\right]}\prod_{k=1}^n(z-e^{\pm(2k-1)\pi i/n})=z^n+1.$$

After $(z-e^{it})(z-e^{-it})=z^2-2z\cos t+1$, the product is $$\prod_{k=1}^n\left(z^2-2z\cos[(2k-1)\pi/n]+1\right)=(z^n+1)^2.$$ Taking the logarithmic derivative (w.r.t. $z$), we obtain $$\sum_{k=1}^n\frac{z-\cos[(2k-1)\pi/n]}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{nz^{n-1}}{z^n+1}.$$

To get the claimed equality, multiply by $2z$, subtract $n$, and divide by $z^2-1$.

The computations could be continued with a closed form of Chebyshev polynomials.

An (essentially equivalent) "algebraic" alternative is as follows. \begin{align*} \sum_{k=1}^{2024}\frac{2023-2022\cos\frac{(2k-1)\pi}{2024}}{2021-2020\cos\frac{(2k-1)\pi}{2024}}&=2024\left(1+\frac1{1010}\right)-\frac1{2\cdot 1010^2}f_{2024}\left(\frac{2021}{2020}\right),\\ f_n(z)&:=\sum_{k=1}^n\frac1{z-\cos[(2k-1)\pi/n]}=g_n\left(z\pm\sqrt{z^2-1}\right),\\ g_n(z)&:=\sum_{k=1}^n\frac{2z}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{2nz}{z^2-1}\frac{z^n-1}{z^n+1}, \end{align*} and the last equality is obtained from the factorization $$\color{gray}{\left[\prod_{k=1}^n(z-e^{\pm 2k\pi i/n})=z^n-1\implies\right]}\prod_{k=1}^n(z-e^{\pm(2k-1)\pi i/n})=z^n+1.$$

After $(z-e^{it})(z-e^{-it})=z^2-2z\cos t+1$, the product is $$\prod_{k=1}^n\left(z^2-2z\cos[(2k-1)\pi/n]+1\right)=(z^n+1)^2.$$ Taking the logarithmic derivative (w.r.t. $z$), we obtain $$\sum_{k=1}^n\frac{z-\cos[(2k-1)\pi/n]}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{nz^{n-1}}{z^n+1}.$$

To get the claimed equality, multiply by $2z$, subtract $n$, and multiply by $\frac{2z}{z^2-1}$.