Example 4 Find $\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}$.
Analysis: Considering the use of identity (III), we can obtain
$$
\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \text {. }
$$

Let $l=2 n-k-1$, and combining with equation (III) again yields $\frac{l+1}{\mathrm{C}_{2 n}^{l}}$, creating a cyclic phenomenon, which leads to the solution of the problem.
Let the original expression be $y_{n}$, from the analysis we have
$$
y_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{C_{2 n}^{2 n-k-1}} \text {. }
$$

By making a change of the summation index, set $l=2 n-$ $k-1$, when $k$ varies from 0 to $2 n-1$, $l$ varies from $2 n-1$ to 0. Thus,
$$
\begin{array}{l}
y_{n}=\sum_{l=0}^{2 n-1}(-1)^{2 n-l} \frac{l+1}{\mathbf{C}_{2 n}^{l}} \\
=\sum_{l=0}^{2 n-1}(-1)^{l} \frac{l+1}{\mathbf{C}_{2 n}^{l}}=-y_{n} .
\end{array}
$$

Therefore, $y_{n}=0$, that is,
$$
\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}=0 .
$$