9. Let $x_{1}, x_{2}, \cdots, x_{n}$ take values 7 or -7, and satisfy
(1) $x_{1}+x_{2}+\cdots+x_{n}=0$;
(2) $x_{1}+2 x_{2}+\cdots+n x_{n}=2009$.

Determine the minimum value of $n$.
9. First, when $n=34$,
$$
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{16}=x_{18}=-7, \\
x_{17}=x_{19}=x_{20}=\cdots=x_{34}=7
\end{array}
$$

satisfies the conditions of the problem.
Second, divide both sides of condition (2) by 7, and let $y_{i}=\frac{x_{i}}{7}(i=1,2, \cdots, n)$. Then
$$
\begin{array}{l}
y_{i}= \pm 1, \\
y_{1}+y_{2}+\cdots+y_{n}=0, \\
y_{1}+2 y_{2}+\cdots+n y_{n}=287 .
\end{array}
$$

From equation (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 equation (2), we get
$$
\begin{array}{r}
287 \leqslant(2 k+2 k-1+\cdots+k+1)- \\
\quad(k+k-1+\cdots+2+1)=k^{2} .
\end{array}
$$

Therefore, $k^{2} \geqslant 287 \Rightarrow k \geqslant 17 \Rightarrow n \geqslant 34$.
Thus, the minimum value of $n$ is 34.