2. 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.
2. All possible values of $f(2007)$ are $1,2, \cdots$, 2008.
For any positive integers $m, n$, if $m>n$, then
$$
f(m)=f(n+(m-n)) \geqslant f(n)+f(f(m-n))-1 \geqslant f(n).
$$

Therefore, $f$ is non-decreasing.
For any positive integer $n, f(n) \equiv 1$, clearly satisfies the condition.

Assume $f(n) \neq \equiv 1$, then there exists the smallest positive integer $a$ such that $f(a)>1$.
Thus, for all positive integers $b \geqslant a$, we have
$$
f(b) \geqslant f(a)>1.
$$

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

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 $2 k+c \geqslant f(2 k)=f(k+k)$
$\geqslant f(k)+f(f(k))-1$
$\geqslant f(k)+f(k)-1$
$$
=2(k+c)-1=2 k+(2 c-1).
$$

Thus, $c \leqslant 1$.
Therefore, for all positive integers $n$, we have
$f(n) \leqslant n+1$.
In particular, $f(2007) \leqslant 2008$.
Below, we use examples to prove: $f(2007)$ can take values from 1 to 2008.
Let $f_{j}(n)=\max \{1, n+j-2007\}(j=1$,
$2, \cdots, 2007)$,
$$
\begin{array}{l}
f_{2008}(n)=\left\{\begin{array}{ll}
n, & 2007 \nmid n ; \\
n+1, & 2007 \mid n.
\end{array}\right. \\
\text { Then } f_{j}(2007)=j(j=1,2, \cdots, 2007), \\
f_{2008}(2007)=2008. \\
\text { For } j=1,2, \cdots, 2007, \text { since } f_{j} \text { is non-decreasing, } \\
\text { and } f_{j}(n) \leqslant n, \text { thus, for all positive } \\
\text { integers } n, \text { we have } f_{j}\left(f_{j}(n)\right) \leqslant f_{j}(n) \leqslant n. \\
\text { If } f_{j}(m)=1, \text { then } \\
f_{j}(m+n) \geqslant f_{j}(n) \geqslant f_{j}\left(f_{j}(n)\right) \\
=f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1,
\end{array}
$$

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 $f_{j}\left(f_{j}(n)\right) \leqslant f_{j}(n) \leqslant n$.
If $f_{j}(m)=1$, then
$$
\begin{array}{l}
f_{j}(m+n) \geqslant f_{j}(n) \geqslant f_{j}\left(f_{j}(n)\right) \\
=f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1,
\end{array}
$$

satisfies the condition.
If $f_{j}(m) \neq 1$, then
$f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1$
$\leqslant m+j-2007+n$
$$
=(m+n)+j-2007=f_{j}(m+n).
$$

Therefore, $f_{j}(j=1,2, \cdots, 2007)$ satisfies the condition.
Since for all positive integers $n$,
$$
n+1 \geqslant f_{2008}(n) \geqslant n,
$$

so, $n+1 \geqslant f_{2008}\left(f_{2008}(n)\right)$.
Actually, if $f_{2008}(n)=n$, then
$f_{2008}\left(f_{2008}(n)\right)=n$.
If $f_{2008}(n)=n+1$, then by $2007 \mid n$, we get $2007 \times(n+1)$. Thus,
$f_{2008}\left(f_{2008}(n)\right)=f_{2008}(n+1)=n+1$.
If $2007 \mid (m+n)$, then
$f_{2 \operatorname{Cos}}(m+n)=m+n+1$
$=(m+1)+(n+1)-1$
$\geqslant f_{2008}(m)+f_{2008}\left(f_{2008}(n)\right)-1$.
If $2007 \mid (m+n)$, then either $2007 \mid m$, or $2007 \mid n$. For the former case, $f_{2008}(m)=$ $m$, and for the latter case, $f_{2008}\left(f_{2008}(n)\right)=f_{2008}(n)$ $=n$. Hence
$$
\begin{array}{l}
f_{2008}(m)+f_{2008}\left(f_{2008}(n)\right)-1 \\
\leqslant(m+n+1)-1=f_{2008}(m+n).
\end{array}
$$

Therefore, $f_{2 \cos }$ also satisfies the condition.