Example 2 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}, \sum_{i=1}^{n} \frac{1}{x_{i}}=A$ (constant), for a given positive integer $k$, find the maximum value of $\sum \frac{1}{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}}$. The summation is over all $k-$ tuples $\left(i_{1}, i_{2}, \cdots, i_{k}\right)$ from $1,2, \cdots, n$.
Analysis and Solution The difficulty of this problem lies in the complexity of the objective function, with the expectation to simplify it using inequalities. Given the structural characteristics of the objective function, the idea is to transform $\frac{1}{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}}$ into $\frac{1}{x_{i_{1}}}+\frac{1}{x_{i_{2}}}+\cdots+\frac{1}{x_{i_{k}}}$ to utilize the condition $\sum_{i=1}^{n} \frac{1}{x_{i}}=A$. This precisely matches the characteristics of the "reciprocal inequality": $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}} \geqslant \frac{k^{2}}{a_{1}+a_{2}+\cdots+a_{k}}$. Therefore, using the "reciprocal inequality," we have
$$\frac{1}{a_{1}+a_{2}+\cdots+a_{k}} \leqslant \frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}}{k^{2}}$$

Thus,
$$\begin{aligned}
\sum \frac{1}{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}} & \leqslant \sum \frac{\frac{1}{x_{i_{1}}}+\frac{1}{x_{i_{2}}}+\cdots+\frac{1}{x_{i_{k}}}}{k^{2}} \\
& =\frac{1}{k^{2}} \sum\left(\frac{1}{x_{i_{1}}}+\frac{1}{x_{i_{2}}}+\cdots+\frac{1}{x_{i_{k}}}\right)
\end{aligned}$$

Consider the number of times each term $\frac{1}{x_{i_{j}}}(j=1,2, \cdots, k)$ appears in the "sum" on the right side of the above equation. Clearly, $\frac{1}{x_{i_{j}}}$ appears once, which is equivalent to the appearance of a $\left(\frac{1}{x_{1}}, \frac{1}{x_{2}}, \cdots, \frac{1}{x_{k}}\right)$ combination containing $\frac{1}{x_{i_{j}}}$. Since there are $\mathrm{C}_{n-1}^{k-1}$ $k$-combinations containing $\frac{1}{x_{i_{j}}}$, $\frac{1}{x_{i_{j}}}$ appears $\mathrm{C}_{n-1}^{k-1}$ times in the "sum," so
$$\frac{1}{k^{2}} \sum\left(\frac{1}{x_{i_{1}}}+\frac{1}{x_{i_{2}}}+\cdots+\frac{1}{x_{i_{k}}}\right)=\frac{1}{k^{2}} \mathrm{C}_{n-1}^{k-1} \sum_{i=1}^{n} \frac{1}{x_{i}}=\frac{A}{k^{2}} \mathrm{C}_{n-1}^{k-1} .$$

Equality holds when $x_{1}=x_{2}=\cdots=x_{n}=\frac{n}{A}$.
Therefore, the maximum value of $\sum \frac{1}{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}}$ is $\frac{A}{k^{2}} \mathrm{C}_{n-1}^{k-1}$.