Three. (50 points) Given $n$ as a positive integer no less than 2. For all non-negative arrays $a_{i}(i=1,2, \cdots, n)$, find the maximum value of $\frac{\left(\sum_{i=1}^{n} i a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{i}\right)}{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}$.
Three, by homogeneity, we may assume $\sum_{i=1}^{n} a_{i}=1$.
We seek to find the maximum value of $\left(\sum_{i=1}^{n} i a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{i}\right)$.
Notice,
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} i a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{i}\right)=\frac{\left(\sum_{i=1}^{n} i a_{i}\right)\left(\sum_{i=1}^{n} \frac{n a_{i}}{i}\right)}{n} \\
\leqslant \frac{\left(\frac{\sum_{i=1}^{n} i a_{i}+\sum_{i=1}^{n} \frac{n a_{i}}{i}}{2}\right)^{2}}{n} . \\
=\frac{\left[\sum_{i=1}^{n} a_{i}\left(i+\frac{n}{i}\right)\right]^{2}}{4 n} .
\end{array}
$$

Given $i \in[1, n]$, we know
$$
\begin{array}{l}
i+\frac{n}{i} \leqslant \max \left\{1+\frac{n}{1}, n+\frac{n}{n}\right\}=n+1 . \\
\text { Hence } \frac{\left[\sum_{i=1}^{n} a_{i}\left(i+\frac{n}{i}\right)\right]^{2}}{4 n} \\
\leqslant \frac{(n+1)^{2}\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{4 n}=\frac{(n+1)^{2}}{4 n} .
\end{array}
$$

When $a_{1}=a_{n}=\frac{1}{2}, a_{2}=a_{3}=\cdots=a_{n-1}=0$,
the equality holds.
Therefore, the maximum value of $\frac{\left(\sum_{i=1}^{n} i a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{i}\right)}{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}$ is $\frac{(n+1)^{2}}{4 n}$.