Three. (50 points) Let $n$ be a given positive integer greater than 1. For any $d_{1}, d_{2}, \cdots, d_{n}>1$, when
$$
\sum_{1 \leqslant i<j \leqslant n} d_{i} d_{j}=(n-1) \sum_{i=1}^{n} d_{i}
$$

find the maximum value of
$$
S=\sum_{i=1}^{n}\left[d_{i} \sum_{j=1}^{n} d_{j}-n d_{i}+n(n-2)\right]^{-1}
$$
$$
\text { Three, } S_{\max }=\frac{1}{n} \text {. }
$$

First, let $d_{1}=d_{2}=\cdots=d_{n}=2$. Clearly, $\sum_{1 \leqslant i0(i=1,2, \cdots, n)$, so,
$$
\sum_{i=1}^{n} a_{i} \geqslant n \sqrt{\frac{\sum_{1 \leq i<j \leq n} a_{i} a_{j}}{C_{n}^{2}}}=n \text {. }
$$

Then, $S=\sum_{i=1}^{n}\left[d_{i} \sum_{j=1}^{n} d_{j}-n d_{i}+n(n-2)\right]^{-1}$
$$
\begin{array}{l}
=\sum_{i=1}^{n}\left[d_{i} \sum_{j=1}^{n}\left(d_{j}-1\right)+n(n-2)\right]^{-1} \\
=\sum_{i=1}^{n}\left[\left(a_{i}+1\right) \sum_{j=1}^{n} a_{j}+n(n-2)\right]^{-1} \\
\leqslant \sum_{i=1}^{n}\left[a_{i} \sum_{j=1}^{n} a_{j}+n(n-1)\right]^{-1} \\
=\sum_{i=1}^{n} \frac{2 \sum_{1 \leq s<1 \leqslant n} a_{s} a_{t}}{n(n-1)\left(a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1 \leqslant s<1 \leqslant n} a_{s} a_{t}\right)} \\
=\frac{1}{(n-1)}\left[1-\frac{1}{n} \sum_{i=1}^{n} \frac{a_{i} \sum_{j=1}^{n} a_{j}}{a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1 \leqslant s<1 \leqslant n} a_{s} a_{t}}\right] .
\end{array}
$$

To prove that (1) holds, it is sufficient to prove that
$$
\sum_{i=1}^{n} \frac{a_{i} \sum_{j=1}^{n} a_{j}}{a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1 \leqslant s<1 \leqslant n} a_{s} a_{i}} \geqslant 1 .
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sum_{i=1}^{n}\left[a_{i}\left(a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1 \leq s<1 \leqslant n} a_{s} a_{t}\right)\right] . \\
\sum_{i=1}^{n} \frac{a_{i}}{a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1 \leqslant s<1 \leqslant n} a_{s} a_{t}} \\
\geqslant\left(\sum_{i=1}^{n} a_{i}\right)^{2} .
\end{array}
$$

Notice that
$$
\begin{array}{l}
\sum_{i=1}^{n} a_{i}\left(a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1=s<1 \in n} a_{x} a_{i}\right)=\left(\sum_{i=1}^{n} a_{i}\right)^{3} . \\
\text { Then } \sum_{i=1}^{n} \frac{a_{i}}{a_{i} \sum_{j=1}^{n} a_{j}+2 \sum_{1 \leq 1<1 \leqslant n}^{n} a_{1} a_{i}} \geqslant \frac{1}{\sum_{i=1}^{n} a_{i}} .
\end{array}
$$

Therefore, (2) holds.
Thus, (1) holds.