1. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=2008$, for $n>1$, $a_{1}+a_{2}+\cdots+a_{n}=(1+2+\cdots+n) a_{n}$. Then $a_{2007}=$ $\qquad$
$\begin{array}{l}\text { 2 } 1 \cdot \frac{6}{2009} \cdot \\ \text { Given } a_{1}+a_{2}+\cdots+a_{n}=(1+2+\cdots+n) a_{n} \\ \Rightarrow a_{1}+a_{2}+\cdots+a_{n-1}=(1+2+\cdots+n-1) a_{n-1} \\ \Rightarrow a_{n}=(1+2+\cdots+n) a_{n}-(1+2+\cdots+n-1) a_{n-1} \\ \quad=\frac{n(n+1)}{2} a_{n}-\frac{n(n-1)}{2} a_{n-1} \\ \Rightarrow\left[\frac{n(n+1)}{2}-1\right] a_{n}=\frac{n(n-1)}{2} a_{n-1} \\ \Rightarrow \frac{a_{n}}{a_{n-1}}=\frac{n}{n+2}(n \geqslant 2) \\ \Rightarrow a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \cdot \frac{a_{2}}{a_{1}} \cdot a_{1} \\ \quad=\frac{n}{n+2} \cdot \frac{n-1}{n+1} \cdots \cdots \frac{2}{4} \times 2008=\frac{6 \times 2008}{(n+1)(n+2)} \\ \Rightarrow a_{200}=\frac{6}{2009} .\end{array}$