1. Let $S_1=\sum_{k=1}^{1012}\frac{1}{\alpha+k}$ and $S_2=\frac{1}{2\cdot1}+\frac{1}{4\cdot3}+\frac{1}{6\cdot5}+\cdots+\frac{1}{2024\cdot2023}.$

Given, $S_1-S_2=\frac{1}{2024}.$

2. Simplify $S_2$.

The general term is $\frac{1}{(2n)(2n-1)},\qquad n=1,2,\dots,1012.$ So, $S_2=\sum_{n=1}^{1012}\frac{1}{(2n)(2n-1)}.$

Use partial fractions: $\frac{1}{(2n)(2n-1)}=\frac{1}{2n-1}-\frac{1}{2n}.$ Hence, $S_2=\sum_{n=1}^{1012}\left(\frac{1}{2n-1}-\frac{1}{2n}\right).$

3. Rewrite the given equation: $\sum_{k=1}^{1012}\frac{1}{\alpha+k}=S_2+\frac{1}{2024}.$

Now observe that $S_2+\frac{1}{2024}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{2023}-\frac{1}{2024}\right)+\frac{1}{2024}.$ So the last two terms cancel, giving $S_2+\frac{1}{2024}=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2023}.$

But a more useful observation is: $\frac{1}{m(m+1)}=\frac{1}{m}-\frac{1}{m+1}.$ If we take $\sum_{m=1012}^{2023}\frac{1}{m(m+1)},$ then it telescopes to $\sum_{m=1012}^{2023}\left(\frac{1}{m}-\frac{1}{m+1}\right)=\frac{1}{1012}-\frac{1}{2024}.$ This is not directly our $S_2+\frac1{2024}$ form, so instead let us search for a telescoping form for $S_1$.

4. Since $S_1$ is a sum of 1012 consecutive reciprocals, $S_1=\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\cdots+\frac{1}{\alpha+1012}.$ We want this to equal $S_2+\frac1{2024}$.

Now compute $S_2+\frac1{2024}$ directly as $\sum_{n=1}^{1012}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\frac1{2024}.$ Since $\frac1{2024}=\frac1{2\cdot 1012+0}$, this suggests matching with the harmonic block $\frac1{1012}+\frac1{1013}+\cdots+\frac1{2023}.$ Let us test $\alpha=1011$.

Then $S_1=\sum_{k=1}^{1012}\frac1{1011+k}=\sum_{m=1012}^{2023}\frac1m.$ So the given equation becomes $\sum_{m=1012}^{2023}\frac1m-S_2=\frac1{2024}.$ Equivalently, S_2=\sum_{m=1012}^{2023}\frac1m-\frac1{2024}=rac1{1012}+\frac1{1013}+\cdots+\frac1{2023}-\frac1{2024}. Now note that this is exactly $\sum_{n=1}^{1012}\left(\frac1{2n-1}-\frac1{2n}\right),$ which is the alternating harmonic block from $1$ to $2024$. This identity is consistent with the standard split of harmonic sums into odd and even parts, and gives precisely the required equality.

Thus $\alpha=1011$ satisfies the equation.

5. Since the sum on the left side $S_1$ is strictly decreasing in $\alpha$, the solution is unique.

Therefore, $\boxed{\alpha=1011}.$