Example 6 Given 2014 real numbers $x_{1}, x_{2}, \cdots, x_{2014}$ satisfy the system of equations
$$
\sum_{k=1}^{2014} \frac{x_{k}}{n+k}=\frac{1}{2 n+1}(n=1,2, \cdots, 2014) .
$$

Try to calculate the value of $\sum_{k=1}^{2014} \frac{x_{k}}{2 k+1}$.
$$
\begin{aligned}
f(x)= & {\left[\prod_{i=1}^{2014}(x+i)\right] . } \\
& {\left[(2 x+1)\left(\sum_{i=1}^{2014} \frac{x_{i}}{x+i}\right)-1\right] . }
\end{aligned}
$$

From the conditions, we know that when $x$ takes the values $1, 2, \cdots, 2014$, we have $f(x)=0$.
Therefore, there exists a constant $c$ such that
$$
f(x)=c(x-1)(x-2) \cdots(x-2014) \text {. }
$$

In equations (1) and (2), taking $x=-\frac{1}{2}$, we get
$$
c=\frac{1}{4029} \text {. }
$$

Thus, $f(x)=\frac{1}{4029} \prod_{i=1}^{2014}(x-i)$.
Hence, $\left[\prod_{i=1}^{2014}(x+i)\right]\left[(2 x+1)\left(\sum_{i=1}^{2014} \frac{x_{i}}{x+i}\right)-1\right]$
$$
=\frac{1}{4029} \prod_{i=1}^{2014}(x-i) \text {. }
$$

Substituting $x=\frac{1}{2}$ in the above equation, we get
$$
\sum_{k=1}^{2014} \frac{x_{k}}{2 k+1}=\frac{1}{4}\left(1-\frac{1}{4029^{2}}\right) .
$$