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