131 Given that the function $f(n)$ is a strictly increasing function defined on $\mathbf{N}_{+}$, with its range also in $\mathbf{N}_{+}$, and satisfies $f(f(n))=3n$. Find $f(2003)$.
Solution: (1) First, $f(1) \neq 1$. Otherwise, on one hand, $f(f(1))=f(1)=1$, and on the other hand, according to the given condition, $f(f(1))=3 \times 1=3$. This is a contradiction.

Second, $f(1) \neq 3$. This contradicts the known condition $f(f(1))=3 \times 1$.
In summary, $f(1)=2$ (since $f(n) \in \mathbf{N}_{+}$).
(2) From $f(f(n))=3 n$, we get
$$
\begin{array}{l}
f(3 n)=f(f(f(n)))=3 f(n). \\
\text { Therefore, } f\left(3^{n}\right)=3 f\left(3^{n-1}\right)=3^{2} f\left(3^{n-2}\right)=\cdots \\
=3^{n} f(1)=2 \times 3^{n}(n \in \mathbf{N}) .
\end{array}
$$
(3) From (2), $f\left(2 \times 3^{n}\right)=f\left(f\left(3^{n}\right)\right)=3 \times 3^{n}$, so,
$$
f\left(2 \times 3^{n}\right)-f\left(3^{n}\right)=3 \times 3^{n}-2 \times 3^{n}=3^{n}.
$$

And $2 \times 3^{n}-3^{n}=3^{n}$, and $f(n+1) \geqslant f(n)+1$ (since $f(n)$ is a strictly increasing function on $\mathbf{N}_{+}$, $\left.f(n) \in \mathbf{N}_{+}\right)$. Therefore,
$$
\begin{array}{l}
f\left(3^{n}+l\right)=f\left(3^{n}\right)+l=2 \times 3^{n}+l\left(l=1,2, \cdots, 3^{n}\right). \\
\text { (4) } f\left(2 \times 3^{n}+l\right)=f\left(f\left(3^{n}+l\right)\right)=3 \times\left(3^{n}+l\right) \\
=3^{n+1}+3 l \quad\left(l=1,2, \cdots, 3^{n}\right).
\end{array}
$$

Thus, $f(2003)=f\left(2 \times 3^{6}+545\right)=3^{7}+3 \times 545=3822$.
(He Bin, Hubei Province Gucheng County Third Senior High School,
441700)