Let $n \in \mathbf{N}_{+}$. Find the positive real number $T(n)$ such that the set of real numbers $x$ satisfying the inequality $\sum_{k=1}^{n} \frac{k}{x-k} \geqslant T(n)$ is the union of disjoint intervals, and the total length of these intervals is 2009.
Consider the function $f(x)=\sum_{k=1}^{n} \frac{k}{x-k}-T(n)$.
It is easy to see that when $k=1,2, \cdots, n$,
$$
\lim _{x \rightarrow k^{+}} f(x)=+\infty, \lim _{x \rightarrow k^{-}} f(x)=-\infty \text {. }
$$

And $f(x)$ is a strictly decreasing continuous function in $(k, k+1)$, therefore, $f(x)=0$ has exactly one real root $x_{k}(k=1,2, \cdots, n-1)$ in $(k, k+1)$.

Also, $\lim _{x \rightarrow \infty} f(x)=-T(n)<0$, and $f(x)$ is a strictly decreasing continuous function in $(-\infty, 1)$ and $(n,+\infty)$, thus, $f(x)=0$ has no real roots in $(-\infty, 1)$ and has a unique real root $x_{n}$ in $(n,+\infty)$.
Consider the equation
$$
f(x) \prod_{k=1}^{n}(x-k)=0 \text {. }
$$

Clearly, $x_{k}(k=1,2, \cdots, n)$ are the real roots of equation (1).
Since the left side of equation (1) is an $n$-degree polynomial, it has at most $n$ real roots. This indicates that $x_{k}(k=1,2, \cdots, n)$ are the unique real roots of $f(x)=0$ in the corresponding intervals.

Therefore, the solution set of the inequality $f(x) \geqslant 0$ is $\bigcup_{k=1}^{n}\left(k, x_{k}\right]$, and the sum of the lengths of these intervals is
$$
\sum_{k=1}^{n}\left(x_{k}-k\right)=\sum_{k=1}^{n} x_{k}-\frac{n(n+1)}{2} .
$$

In equation (1), the coefficient of the $n$-degree term is $-T(n)$, and the coefficient of the $(n-1)$-degree term is
$$
\begin{array}{l}
\sum_{k=1}^{n} k-T(n) \sum_{k=1}^{n}(-k) \\
=(1+T(n)) \frac{n(n+1)}{2},
\end{array}
$$

Thus, $\sum_{k=1}^{n} x_{k}=\frac{n(n+1)}{2} \cdot \frac{1+T(n)}{T(n)}$.
Therefore, the sum of the lengths of all intervals is
$$
\sum_{k=1}^{n}\left(x_{k}-k\right)=\frac{n(n+1)}{2 T(n)}=2009 \text {. }
$$

Hence, $T(n)=\frac{n(n+1)}{4018}$.