Let $n$ be a positive integer, and let $f(n)$ denote the number of positive integers not exceeding $\sqrt{n}$ (for example, $f(3)=1$, $f(9)=3$).
(1) Find $f(2007)$;
(2) Find the positive integer $n$ such that it satisfies
$$
f(1)+f(2)+\cdots+f(n)=2009 .
$$
(1) From $44<\sqrt{2007}<45$, we have $f(2007)=44$.
(2) When $k=1,2,3$, $f(k)=1$, then
$$
f(1)+f(2)+f(3)=1 \times 3 \text {. }
$$

When $k=4,5, \cdots, 8$, $f(k)=2$, then
$$
f(4)+f(5)+\cdots+f(8)=2 \times 5 \text {. }
$$

When $k=9,10, \cdots, 15$, $f(k)=3$, then
$$
f(9)+f(10)+\cdots+f(15)=3 \times 7 \text {. }
$$

When $k=16,17, \cdots, 24$, $f(k)=4$, then
$$
f(16)+f(17)+\cdots+f(24)=4 \times 9 \text {. }
$$

When $k=169,170, \cdots, 195$, $f(k)=13$, then
$$
f(169)+f(170)+\cdots+f(195)=13 \times 27 \text {. }
$$

When $k=196,197, \cdots, 224$, $f(k)=14$.
From the original equation, we have
$$
\begin{array}{l}
1 \times 3+2 \times 5+3 \times 7+4 \times 9+\cdots+13 \times 27+14 x \\
=2009,
\end{array}
$$

Solving for $x$ gives $x=20$.
Thus, $n=195+20=215$, which means
$$
f(1)+f(2)+\cdots+f(215)=2009 \text {. }
$$

Therefore, the natural number $n=215$.