13) (10 points) If there exist $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+2 x_{2}+\cdots+$ $n x_{n}=2009$, and $x_{1}+x_{2}+\cdots+x_{n}=0$, where $x_{i}= \pm 7, i=1,2, \cdots$,
$n$. Determine the minimum value of $n$.
13 Notice that $2009=7 \times 287$. Let $y_{i}=\frac{x_{i}}{7}, i=1,2, \cdots, n$, then $y_{i}=$ $\pm 1$
$$
\begin{aligned}
y_{1}+y_{2}+\cdots+y_{n} & =0, \\
y_{1}+2 y_{2}+\cdots+n y_{n} & =287 .
\end{aligned}
$$

From (1), we know that $n$ is even. Let $n=2 k$, then $y_{1}, y_{2}, \cdots, y_{n}$ contain $k$ -1s and $k$ 1s.
From (2), we get
$$
\begin{aligned}
287 & \leqslant(2 k+(2 k-1)+\cdots+(k+1))-(k+(k-1)+\cdots+2+1) \\
& =k^{2} .
\end{aligned}
$$

Therefore, $k^{2} \geqslant 287$, which means $k \geqslant 17, n \geqslant 34$.
When $n=34$, take
$$
\begin{array}{l}
y_{1}=y_{2}=\cdots=y_{16}=y_{18}=-1, \\
y_{17}=y_{19}=y_{20}=\cdots=y_{34}=1,
\end{array}
$$

then (1) and (2) are satisfied. Therefore, the minimum value of $n$ is 34.