Example 15 (48th IM() Preliminary Question) For all $m, n \in \mathbf{N}^{+}$, the function $f: \mathbf{N}^{+} \rightarrow \mathbf{N}^{+}$ satisfies
$$
f(m+n) \geqslant f(m)+f[f(n)]-1 .
$$

Find all possible values of $f(2007)$, where $\mathbf{N}^{+}$ is the set of positive integers.
Solve for all possible values of $f(2007)$ as $1,2, \cdots, 2008$.
For any positive integers $m, n$, if $m>n$, then
$$
\begin{aligned}
f(m) & =f[n+(m-n)] \\
& \geqslant f(n)+f[f(m-n)]-1 \geqslant f(n) .
\end{aligned}
$$

Therefore, $f$ is non-decreasing.
For any positive integer $n, f(n)=1$, it clearly satisfies the condition.
Assume $f(n) \neq 1$, then there exists the smallest positive integer $a$, such that $f(a)>1$.
Thus, for all positive integers $\dot{o} \geqslant a$, we have
$$
f(b) \geqslant f(a)>1 .
$$

If there exists a positive integer $n$, such that $f(n)>n$, then
$$
\begin{aligned}
f[f(n)] & =f\{[f(n)-n]+n\} \\
& \geqslant f[f(n)-n]+f[f(n)]-1 .
\end{aligned}
$$

Therefore, $f[f(n)-n] \leqslant 1$.
Thus, $f(n)-n<a$.
Let $f(n)-n$ take the maximum value $c$ when $n=k$.
Then $f(k)-k=c \geqslant 1$.
Hence $2 k+c \geqslant f(2 k)=f(k+k)$
$$
\begin{array}{l}
\geqslant f(k)+f[f(k)]-1 \\
\geqslant f(k)+f(k)-1 \\
=2(k+c)-1=2 k+(2 c-1) .
\end{array}
$$

Thus, $c \leqslant 1$.
Therefore, for all positive integers $n$, we have
$$
f(n) \leqslant n+1 \text {. }
$$

In particular, $f(2007) \leqslant 2008$.
Below, we use examples to prove: $f(2007)$ can take values from 1 to 2008.
Let $f,(n)=\max \{1, n+j-2007\}(j=1,2, \cdots, 2007)$,
$$
f_{\text {zoons }}(n)=\left\{\begin{array}{ll}
n, & 2007 \times n ; \\
n+1, & 2007 \mid n .
\end{array}\right.
$$

Then $f_{i}(2007)=j(j=1,2, \cdots, 2007)$,
$$
f_{\text {zoum: }}(2007)=2008 \text {. }
$$

For $j=1,2, \cdots, 2007$, since $f_{j}$ is non-decreasing, and $f_{j}(n) \leqslant n$, thus, for all positive integers $n$, we have $f,\left[f_{i}(n)\right] \leqslant f_{j}(n) \leqslant n$.
If $f,(m)=1$, then
$$
\begin{aligned}
f_{1}(m+n) & \geqslant f_{j}(n) \geqslant f_{j}\left[f_{j}(n)\right] \\
& =f_{i}(m)+f_{j}\left[f_{j}(n)\right]-1 .
\end{aligned}
$$

It satisfies the condition.
If $f,(m) \neq 1$, then
$$
\begin{array}{l}
f_{1}(m)+f_{1}\left[f_{1}(n)\right]-1 \\
\leqslant m+j-2007+n \\
=(m+n)+j-2007=f_{1}(m+n) .
\end{array}
$$

Therefore, $f_{j}(j=1,2, \cdots, 2 \operatorname{co7})$ satisfies the condition.
Since for all positive integers $n$,
$$
n+1 \geqslant f_{\text {200x }}(n) \geqslant n,
$$

Thus, $n+1 \geqslant f_{2 \text { mous }}\left[f_{\text {son.s }}(n)\right]$.
Actually, if $f_{\text {moox }}(n)=n$, then
$$
f_{2 \operatorname{mons}}\left[f_{2 \text { moox }}(n)\right]=n \text {. }
$$

If $f_{\text {zmumx }}(n)=n+1$. Then by $2007 \mid n$, we get $2007 \times(n+1)$. Thus,
$$
f_{\operatorname{monx}}\left[f_{\operatorname{2ong}}(n)\right]=f_{200 x}(n+1)=n+1 .
$$

If $2007 \mid(m+n)$, then
$$
\begin{aligned}
f_{\text {2008 }}(m+n) & =m+n+1 \\
& =(m+1)+(n+1)-1 \\
& \geqslant f_{2008}(m)+f_{2 \text { mons }}\left[f_{2 \text { mos }}(n)\right]-1 .
\end{aligned}
$$

If $2007 \times(m+n)$, then either $2007 \times m$, or $2007 \times n$, for the former case we have $f_{20 o s}(m)=m$, for the latter case we have $f_{5 \operatorname{mos} x}\left[f_{2 \operatorname{mox}}(n)\right]=f_{2008}(n)=n$. Hence
s.
Therefore, $f_{200 x}$ also satisfies the condition.