Five. (20 points) Let 2006 real numbers $a_{1}, a_{2}, \cdots, a_{2006}$ satisfy
$$
\begin{array}{l}
\frac{a_{1}}{2}+\frac{a_{2}}{3}+\cdots+\frac{a_{2006}}{2007}=\frac{4}{3}, \\
\frac{a_{1}}{3}+\frac{a_{2}}{4}+\cdots+\frac{a_{2006}}{2008}=\frac{4}{5}, \\
\frac{a_{1}}{4}+\frac{a_{2}}{5}+\cdots+\frac{a_{2006}}{2009}=\frac{4}{7}, \\
\cdots \cdots \\
\frac{a_{1}}{2007}+\frac{a_{2}}{2008}+\cdots+\frac{a_{2006}}{4012}=\frac{4}{4013} .
\end{array}
$$

Find the value of the algebraic expression
$$
\frac{a_{1}}{3}+\frac{a_{2}}{5}+\frac{a_{3}}{7}+\cdots+\frac{a_{2006}}{4013}
$$
Let $n=2006$, and define
$$
R(x)=\frac{a_{1}}{x+1}+\frac{a_{2}}{x+2}+\cdots+\frac{a_{n}}{x+n} .
$$

Then the required algebraic expression is
$$
\frac{a_{1}}{3}+\frac{a_{2}}{5}+\frac{a_{3}}{7}+\cdots+\frac{a_{2006}}{4013}=\frac{1}{2} R\left(\frac{1}{2}\right) \text {. }
$$

Let $R(x)=\frac{q(x)}{p(x)}$,
where $p(x)=(x+1)(x+2) \cdots(x+n)$, and $q(x)$ is a polynomial of degree less than $n$.
From the definition, we know
$$
R(k)=\frac{4}{2 k+1}(\forall k=1,2, \cdots, n) .
$$

Thus, the polynomial $4 p(x)-(2 x+1) q(x)$ has a degree no greater than $n$, and $1,2, \cdots, n$ are roots of the equation
$$
4 p(x)-(2 x+1) q(x)=0
$$

By the fundamental theorem of algebra, we have
$$
\begin{array}{l}
4 p(x)-(2 x+1) q(x) \\
=c(x-1)(x-2) \cdots(x-n) .
\end{array}
$$

where $c$ is a constant dependent on $a_{1}, a_{2}, \cdots, a_{n}$. By comparing the coefficients of $x^{n}$ on both sides of equation (3), we get
$$
c=4-2\left(a_{1}+a_{2}+\cdots+a_{n}\right) .
$$

Substituting $x=-\frac{1}{2}$ into equation (3), we get
$$
c=\frac{4(-1)^{n}}{2 n+1} \text {. }
$$

Additionally, substituting $x=\frac{1}{2}$ into equation (3) yields
$$
4 p\left(\frac{1}{2}\right)-2 q\left(\frac{1}{2}\right)=\frac{4(1 \times 3 \times 5 \times \cdots \times(2 n-1))}{(2 n+1) \times 2^{n}} \text {. }
$$

From equation (2), we have
$$
R\left(\frac{1}{2}\right)=\frac{q\left(\frac{1}{2}\right)}{p\left(\frac{1}{2}\right)}=2-\frac{2}{(2 n+1)^{2}} \text {. }
$$

Substituting the above equation into equation (1), we get
$$
\frac{1}{2} R\left(\frac{1}{2}\right)=1-\frac{1}{(2 n+1)^{2}}=1-\frac{1}{4013^{2}} \text {. }
$$