Let $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ be a function, and for any positive integers $m, n$, we have
$$
f(f(m)+f(n))=m+n .
$$

Find the value of $f(2011)$.
Since $f(x)$ is a function from the set of positive integers to the set of positive integers, let
$$
f(1)=p\left(p \in \mathbf{N}_{+}\right) \text {. }
$$

If $p>1$, then $p \geqslant 2$.
$$
\text { Let } p=1+b\left(b \in \mathbf{N}_{+}\right), f(b)=c\left(c \in \mathbf{N}_{+}\right) \text {. }
$$

On one hand,
$$
\begin{array}{l}
f(f(1)+f(1))=f(p+p) \\
=f(1+b+1+b)=f(2+2 b) \\
=f(f(f(1)+f(1))+f(f(b)+f(b))) \\
=f(f(2 f(1))+f(2 c))=2 f(1)+2 c \\
=2 p+2 c=2+2 b+2 c>2 .
\end{array}
$$

On the other hand, $f(f(1)+f(1))=2$. This is a contradiction.
Therefore, $p=1$.
Thus, $f(1)=1$,
$$
f(2)=f(1+1)=f(f(1)+f(1))=2 \text {. }
$$

Conjecture: $f(n)=n$.
We will prove this by mathematical induction.
When $n=1$, $f(1)=1$, so the conclusion holds.
Assume that when $n=k$, the conclusion holds.
When $n=k+1$,
$$
f(k+1)=f(f(k)+f(1))=k+1 \text {. }
$$

Thus, when $n=k+1$, the conclusion holds.
Therefore, for all $n \in \mathbf{N}_{+}, f(n)=n$.
Hence, $f(2011)=2011$.
(Pan Tie, Tianjin Experimental High School, 300074)