$$ \sum_{r = 1}^{2023} \frac{\,\left(-1\right)^{r - 1}r\,}{\operatorname{C}_{r}^{2024}} $$
My approach was to add up $1_{\rm st}$ and $2023_{\rm rd}$ term, $2_{\rm nd}$ and $2022_{\rm nd}$, $3_{\rm rd}$ and $2021_{\rm st}$ and so on $$ T_r + T_{2024 - r} = \frac{2024\left(-1\right)^{r - 1}}{\operatorname{C}_{r}^{2024}} $$ Cant think of anything beyond this ?.
$$\begin{align} \sum_{r=1}^{2023}\frac{(-1)^{r+1}\cdot r}{2024\choose r}&=\sum_{r=1}^{2023}\frac{(-1)^{r+1}((r+1)-1)}{2024\choose r}\\ &=\sum_{r=1}^{2023}\frac{(-1)^{r+1}(r+1)}{2024\choose r}-\sum_{r=1}^{2023}\frac{(-1)^{r+1}}{2024\choose r}\\ &=2025\sum_{r=1}^{2023}\frac{(-1)^{r+1}}{2025\choose r+1}-\sum_{r=1}^{2023}\frac{(-1)^{r+1}}{2024\choose r}\\ &=2025\cdot\frac{1}{2025 \choose 2024}-\sum_{r=1}^{2023}\frac{(-1)^{r+1}}{2024\choose r}\tag{a}\\ &=2025\cdot\frac{1}{2025 \choose 2024}-\frac{1}{1013}\tag{b}\\ &=\frac{1012}{1013} \end{align}$$
Explanations:
$\text{(a): Cancelling out equal terms in the first summation}$
$\text{(b): Using }$MSE 1402886$\text{ in the other summation}$
Inspired from here,
$$S=\sum_{r=1}^{2023}\frac{(-1)^{r-1}r}{\binom{2024}{r}}=2025\sum_{r=1}^{2023}\frac{(-1)^{r-1}r}{2025\binom{2024}{r}}$$ $$=2025\sum_{r=1}^{2023}\frac{(-1)^{k-1}r}{2025\binom{2024+r-r}{r}}$$
Note,
$$\frac{1}{2025\binom{2024+r-r}{r}}=\int_0^1 x^r (1-x)^{2024-r} \,dx$$
Or in general from here, beta functions for the win,
$$\boxed{\int_0^1 x^a(1-x)^b=\frac{1}{\binom{a+b}{a}(1+a+b)}}$$
$$=2025\sum_{r=1}^{2023}{(-1)^{r-1}(r)} \int_0^1 x^r (1-x)^{2024-r} \,dx$$ $$=2025\int_0^1\sum_{r=1}^{2023}(-1)^{r-1}(r) x^r (1-x)^{2024-r} \,dx$$ Separating the $r$-dependent terms and taking a negative out, $$=2025\int_0^1 (-1)(x-1)^{2024}\sum_{r=1}^{2023}(r)(x^r)(x-1)^{-r}\,dx$$
$$=2025\int_0^1 (x - 1) (x^{2024} (x - 2024) - x(x-1)^{2024}) \,dx$$
$$=2025\int_0^1 x(x - 1) (x^{2024} - (x-1)^{2024}) \,dx-(2024)(2025)\int_0^1 (x - 1) x^{2024}\,dx$$
Using this, we can eliminate the first integral,
$$=-(2024)(2025)\int_0^1 (x - 1) x^{2024}\,dx=(2024)(2025)\int_0^1 (1-x) x^{2024}\,dx$$
Referring above reference which is boxed, we get,
$$S=\frac{1012}{1013}$$
Found the generalized solution from my old notes,
$$\boxed{S=\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}k}{\binom{2n}{k}}=\frac{n}{n+1}}$$
Seeking to find a closed form of
$$\sum_{k=1}^{2n-1} (-1)^{k-1} k {2n\choose k}^{-1}$$
Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1} = [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$
We can re-write this as
$$\frac{1}{n} {n-1\choose k-1}^{-1} = [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$
Using ${2n\choose 2n-k}$ we obtain
$$(2n+1) [v^{2n+1}] \log\frac{1}{1-v} \sum_{k=1}^{2n-1} (-1)^{k-1} k (v-1)^k \\ = (2n+1) [v^{2n+1}] \log\frac{1}{1-v} [z^{2n-1}] \frac{1}{1-z} \sum_{k\ge 1} (-1)^{k-1} k (v-1)^k z^k \\ = (2n+1) [v^{2n+1}] \log\frac{1}{1-v} [z^{2n-1}] \frac{1}{1-z} \frac{(v-1)z}{(1+(v-1)z)^2}.$$
The contribution from $z$ is
$$\;\underset{z}{\mathrm{res}}\; \frac{1}{z^{2n}} \frac{1}{1-z} \frac{(v-1)z}{(1+(v-1)z)^2} \\ = \frac{1}{v-1} \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{2n-1}} \frac{1}{1-z} \frac{1}{(z+1/(v-1))^2}.$$
Here the residue at infinity vanishes and residues sum to zero, so we may evaluate with minus the residues at $z=1$ and at $z=-1/(v-1)$, getting first
$$(2n+1)[v^{2n+1}] \log\frac{1}{1-v} \frac{v-1}{v^2} = (2n+1)[v^{2n+3}] \log\frac{1}{1-v} (v-1) \\ = \frac{2n+1}{2n+3} {2n+2\choose 2n+1}^{-1} = \frac{2n+1}{(2n+2)(2n+3)}.$$
and second
$$\left.\left( \frac{1}{z^{2n-1}} \frac{1}{1-z} \right)'\right|_{z=-1/(v-1)} = \left. -\frac{2n-1}{z^{2n}} \frac{1}{1-z} + \frac{1}{z^{2n-1}} \frac{1}{(1-z)^2} \right|_{z=-1/(v-1)}.$$
In terms of $v$,
$$(2n+1)[v^{2n+1}] \log\frac{1}{v-1} \frac{1}{v-1} \\ \times \left[-(2n-1) (v-1)^{2n} (v-1)/v - (v-1)^{2n-1} (v-1)^2/v^2 \right].$$
This gives two pieces, the first is
$$- (2n+1) (2n-1) [v^{2n+2}] \log\frac{1}{v-1} (v-1)^{2n} \\ = - (2n+1) (2n-1) \frac{1}{2n+2} {2n+1\choose 1}^{-1} = - \frac{2n-1}{2n+2}$$
and the second
$$- (2n+1) [v^{2n+3}] \log\frac{1}{v-1} (v-1)^{2n} \\ = - (2n+1) \frac{1}{2n+3} {2n+2\choose 2}^{-1} = - 2 \frac{1}{2n+2} \frac{1}{2n+3}.$$
Adjust the sign and collect everything to get
$$\frac{2n+3}{(2n+2)(2n+3)} + \frac{2n-1}{2n+2} = \frac{2n}{2n+2} = \frac{n}{n+1}.$$
