10. The function $f$ is defined on the set of positive integers, and satisfies $f(1)=2002, f(1)+f(2)+\cdots+f(n)=n^{2} f(n)$ $(n>1)$, then the value of $f(2002)$ is $\qquad$ .
10. $\frac{2}{2003}$ because $f(1)+f(2)+\cdots+f(n-1)+f(n)=n^{2} f(n)$,

so $f(1)+f(2)+\cdots+f(n-1)=(n-1)^{2} f(n-1)$.
(1) - (2). We get $f(n)=n^{2} f(n)-(n-1)^{2} f(n-1)(n>1)$.

Therefore, $f(n)=\frac{n-1}{n+1} f(n-1)=\frac{n-1}{n+1} \cdot \frac{n-2}{n} f(n-2)=\frac{n-1}{n+1} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n-1} f(n-3)$
$$
=\cdots=\frac{n-1}{n+1} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdot \frac{n-4}{n-2} \cdots \cdot \frac{1}{3} f(1)=\frac{2 f(1)}{(n+1) n}=\frac{2 \times 2002}{(n+1) n} .
$$

Thus, $f(2002)=\frac{2}{2003}$. Therefore, the answer should be $\frac{2}{2003}$.