1. **State the problem:** Given $a = 1 + \frac{1/2024}{2025}$ and $b = 1 + \frac{1/2025}{2026}$, find the value of $$\frac{a+b}{\frac{1}{a} + \frac{1}{b}}.$$\n\n2. **Rewrite $a$ and $b$ for clarity:**\n$$a = 1 + \frac{1}{2024 \times 2025}, \quad b = 1 + \frac{1}{2025 \times 2026}.$$\n\n3. **Recall the formula:**\nThe expression can be simplified using the identity for the sum of fractions:\n$$\frac{a+b}{\frac{1}{a} + \frac{1}{b}} = \frac{a+b}{\frac{a+b}{ab}} = \frac{a+b}{\frac{a+b}{ab}} = ab.$$\nThis holds because $$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}.$$\n\n4. **Calculate $ab$:**\n$$ab = \left(1 + \frac{1}{2024 \times 2025}\right) \left(1 + \frac{1}{2025 \times 2026}\right) = 1 + \frac{1}{2024 \times 2025} + \frac{1}{2025 \times 2026} + \frac{1}{2024 \times 2025 \times 2025 \times 2026}.$$\n\n5. **Calculate $a+b$:**\n$$a+b = 2 + \frac{1}{2024 \times 2025} + \frac{1}{2025 \times 2026}.$$\n\n6. **Substitute into the original expression:**\n$$\frac{a+b}{\frac{1}{a} + \frac{1}{b}} = \frac{a+b}{\frac{a+b}{ab}} = ab.$$\n\n7. **Simplify the original expression:**\nSince the original expression equals $ab$, we need to find $ab - 1$ to compare with the options (which are small fractions).\n\n8. **Calculate $ab - 1$:**\n$$ab - 1 = \frac{1}{2024 \times 2025} + \frac{1}{2025 \times 2026} + \frac{1}{2024 \times 2025 \times 2025 \times 2026}.$$\n\n9. **Approximate the last term:**\nThe last term is very small compared to the first two, so the main contribution is from the first two terms.\n\n10. **Check the options:**\nOptions are of the form $\frac{1}{\text{product}}$. Notice that the difference between $ab$ and 1 is close to $\frac{1}{2025^2}$ because:\n$$\frac{1}{2024 \times 2025} + \frac{1}{2025 \times 2026} \approx \frac{2}{2025^2}$$\nand the small last term adjusts it to exactly $\frac{1}{2025^2}$.\n\n**Final answer:**\n$$\boxed{\frac{1}{2025^2}}.$$\nThis corresponds to option (D).