27. Let $0<p \leqslant x_{i} \leqslant q\left(i=1,2, \cdots, n, p, q\right.$ be given constants), try to find the maximum value of $F\left(x_{4}, x_{2}, \cdots\right.$, $\left.x_{n}\right)=\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right)$. $(n=5$ is from the 1977 USA
27. (1) First prove: when $x_{i}(i=1,2, \cdots, n-1)$ are fixed, $x_{n}=p$ (or $\left.q\right), F\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{n}\right)$ is maximized.

In fact, let $u=x_{1}+x_{2}+\cdots+x_{n-1}, v=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n-1}}$,
since $0<p \leqslant x_{i} \leqslant q(i=1,2, \cdots, n)$, we have $p^{2} \leqslant \frac{u}{v} \leqslant q^{2}$, so $p \leqslant \sqrt{\frac{u}{v}} \leqslant q$.
At this point, $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(u+x_{n}\right)\left(v+\frac{1}{x_{n}}\right)=(1+u v)^{-}+\left(\frac{u}{x_{n}}+v x_{n}\right)$,
it is easy to see that $f\left(x_{n}\right)=\frac{u}{x_{n}}+v x_{n}$ is monotonically decreasing on the interval $\left[p,-\sqrt{\frac{u}{v}}\right]$ and monotonically increasing on $\left[\sqrt{\frac{u}{v}}, q\right]$, so $F\left(x_{1}, x_{2}, \cdots, x_{n=1}, p\right)$ or $F\left(x_{1}, x_{2}, \cdots, x_{n=1}, q\right)$ is maximized.
(2) From (1), $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is maximized only when each $x_{n}$ takes the endpoint value $p$ or $q$.

Let $x_{1}, x_{2}, \cdots, x_{n}$ take $p$ for $k$ times, and $q$ for $n-k$ times,
so $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left[k p+(--k) q\left[\cdot\left[\frac{k}{p}+(n-k) \frac{1}{q}\right]=k(n-k)\right.\right.$

From this, we can get:
When $n$ is even, $k=\frac{n}{2}, F\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is maximized, $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{n^{2}}{4}\left(\sqrt{\frac{p}{q}}-\right.$
$$\left.\sqrt{\frac{q}{p}}\right)^{2}+n^{2}$$

When $n$ is odd, $k=\frac{n \pm 1}{2}, F\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is maximized, $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=$ $\frac{n^{2}-1}{4}\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2}+n^{2}$

When $n=5$, $(\alpha+\beta+\gamma+\delta-\varepsilon)\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}+\frac{1}{-}\right) \leqslant 25$ $6\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2}$, from the above proof process, the inequality in the problem holds if and only if $\alpha, \beta, \gamma, \delta, \varepsilon$ have two values the same as one endpoint of the interval $p, q]$, and the other three values the same as the other endpoint of the interval $[p, q]$.