If $$\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots . .+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$$, then $\alpha$ is equal to ___________.

<p>$$\begin{aligned} & \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\ & \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\ & \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) \\ & \quad=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right) \\ & \quad=\left(1+\frac{1}{3}+\frac{1}{5}+\ldots .+\frac{1}{2023}\right) \\ & \quad-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2024}\right) \\ & \quad=\left(1+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1012}\right) \end{aligned}$$</p> <p>$$\begin{aligned} & =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1011}\right) \\ & \frac{-1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1012}\right) \\ & =\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}-\frac{1}{2024} \\ & \Rightarrow \alpha+1012=2023 \\ & \Rightarrow \alpha=1011 \end{aligned}$$</p>