33. The function $f: \mathbf{N} \rightarrow \mathbf{N} (\mathbf{N}$ being the set of non-negative integers), satisfies:
(1) For any non-negative integer $n$, 
$$
f(n+1)>f(n) \text{; }
$$
(2) For any $m, n \in \mathbf{N}$, 
$$
f(n+f(m))=f(n)+m+1 \text{. }
$$

Find the value of $f(2001)$.
(2001, British Mathematical Olympiad (Round 1))
Solution: Let $f(0)=k$, where $k$ is a non-negative integer. From (2), we have
$$
f(n+k)=f(n)+1 \text {. }
$$

If $k=0$, then $f(n)=f(n)+1$, which is a contradiction. Therefore, $k \neq 0$.
According to (1), we have
$$
f(n+k-1) \cdot 1 \text{, if } k>1, \text{ then } n+k-1 \geqslant n+1. \text{ Thus, by (1), we get }
$$
$$
f(n+k-1) \geqslant f(n+1) \geqslant f(n)+1 \text {. }
$$
(2) and (3) are contradictory, so $k=1$ is the only solution.

When $k=1$, equation (1) becomes $f(n+1)=f(n)+1$. This function satisfies conditions (1) and (2), hence the unique solution is $f(2001)=2002$.