3.2.13 * The function $f: \mathbf{N}^{\star} \rightarrow \mathbf{R}$ satisfies: $f(1)=1003$, and for any positive integer $n$, we have
$$
f(1)+f(2)+\cdots+f(n)=n^{2} f(n),
$$

Find the value of $f(2006)$.
Let $a_{n}=f(n)$, then when $n \geqslant 2$, we have $a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, a_{1}+a_{2}+\cdots$ $+a_{n-1}-(n-1)^{2} a_{n-1}$, so $a_{n}-n^{2} a_{n}-(n-1)^{2} a_{n-1}$, which means $a_{n}=\frac{n-1}{n+1} a_{n-1}$. Therefore, $a_{n}=\frac{n-1}{n+1} a_{n-1}=\frac{n-1}{n+1} \cdot \frac{n-2}{n} a_{n-2}=\frac{n-1}{n+1} \cdot \frac{n-2}{n} \cdots \cdot \frac{1}{3} a_{1}-\frac{2 a_{1}}{n(n+1)}$, hence $a_{2006}=\frac{2 \times 1003}{2006 \times 2007}=\frac{1}{2007}$.