Three. (50 points) Given a positive integer $n$, let $n$ real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following $n$ equations:
$$
\sum_{i=1}^{n} \frac{a_{i}}{i+j}=\frac{4}{2 j+1}(j=1,2,3, \cdots, n) .
$$

Determine the value of the sum $S=\sum_{i=1}^{n} \frac{a_{i}}{2 i+1}$ (expressed in the simplest form involving $n$).
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{a_{i}}{x+i}-\frac{4}{2 x+1} \\
=\frac{P(x)}{(2 x+1) \prod_{i=1}^{n}(x+i)}=0 .
\end{array}
$$

From the given, the degree of $P(x)$ does not exceed $n$. For $x=1,2,3, \cdots, n$, it holds, so $P(x)=k \cdot \prod_{i=1}^{n}(x-i)$ (where $k$ is a constant to be determined). Therefore, we have
$$
\sum_{i=1}^{n} \frac{a_{i}}{x+i} \equiv \frac{4}{2 x+1}+\frac{k \cdot \prod_{i=1}^{n}(x-i)}{(2 x+1) \prod_{i=1}^{n}(x+i)},
$$
Multiplying both sides of (1) by $(2 x+1) \prod_{i=1}^{n}(x+i)$ still results in an identity:
$$
\begin{array}{l}
(2 x+1) \sum_{i=1}^{n} \frac{a_{i}}{x+i} \cdot \prod_{i=1}^{n}(x+i) \\
=4 \prod_{i=1}^{n}(x+i)+k \prod_{i=1}^{n}(x-i)(\forall x \in \mathbf{R}),
\end{array}
$$

This is because the degree of (2) does not exceed $n$, and it has $n$ roots at $x=1,2,3, \cdots, n$. Additionally, there are infinitely many roots (since any number except $-1,-2, \cdots,-n,-\frac{1}{2}$ is a root, due to (1) being an identity). Taking $x=-\frac{1}{2}$, we get
$$
\begin{array}{l}
k\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right) \cdots\left(-\frac{1}{2}-n\right) \\
+4\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}+2\right) \cdots\left(-\frac{1}{2}+n\right)=0,
\end{array}
$$

which simplifies to $(-1)^{n} \cdot k \cdot \frac{3}{2} \cdot \frac{5}{2} \cdots \frac{2 n+1}{2}+4 \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \frac{2 n-1}{2}=0$.
Thus, $k=(-1)^{n} \cdot \frac{-4}{2 n+1}$.
Substituting $k$ into equation (1) and taking $x=\frac{1}{2}$, we get
$$
\begin{array}{l}
2 S=\sum_{i=1}^{n} \frac{a_{i}}{i+\frac{1}{2}} \\
=\frac{4}{2}+\frac{k \cdot(-1)^{n} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \cdots \frac{2 n-1}{2}}{2 \cdot \frac{3}{2} \cdot \frac{5}{2} \cdots \cdots \frac{2 n+1}{2}} \\
=2+\frac{(-1)^{n} \cdot k}{2 \cdot(2 n+1)}=2+\frac{\frac{-4}{2 n+1}}{2 \cdot(2 n+1)} \\
=2-\frac{2}{(2 n+1)^{2}},
\end{array}
$$

Therefore, $S=1-\frac{1}{(2 n+1)^{2}}$.