316 Given that the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ is a monotonically increasing function. If $f(f(n))=3 n$, find $f(2011)$.
Let $f(1)=m$. Then $f(m)=f(f(1))=3$.
Thus, $m>1 \Rightarrow f(m)>f(1) \Rightarrow 3>m$.
Therefore, $f(1)=m=2 \Rightarrow f(2)=f(f(1))=3$.
Similarly, $f(3)=6, f(6)=9$.
From $f(3)<f(4)<f(5)<f(6)$, we get
$6<f(4)<f(5)<9$.
Hence, $f(4)=7, f(5)=8$.
Furthermore, $f(7)=f(f(4))=12$,
$f(8)=f(f(5))=15$,
$f(9)=f(f(6))=18$,
$f(12)=f(f(7))=21$,
$f(10)=19, f(11)=20$.
We will prove by mathematical induction:
$$
f(n)=\left\{\begin{array}{ll}
n+3^{m}, & 3^{m} \leqslant n<2 \times 3^{m} ; \\
3 n-3^{m+1}, & 2 \times 3^{m} \leqslant n<3^{m+1}
\end{array}\left(m \in N_{+}\right) .\right.
$$

When $m=0,1$, the conclusion has been verified.
Assume the conclusion holds for $m=k-1$.
When $2 \times 3^{k-1} \leqslant n<3^{k}$,
$f(n)=3 n-3^{k} \Rightarrow f\left(3 n-3^{k}\right)=3 n$.
Let $3 n-3^{k}=p \in\left[3^{k}, 2 \times 3^{k}\right)$. Then
$$
f(p)=p+3^{k} \text {. }
$$

By the monotonicity of $f$, when $3^{k} \leqslant n<2 \times 3^{k}$, we always have $f(n)=n+3^{k}$.
Then $f\left(n+3^{k}\right)=f(f(n))=3 n$.
Let $n+3^{k}=q \in\left[2 \times 3^{k}, 3^{k+1}\right)$. Then
$$
f(q)=3 q-3^{k+1} \text {. }
$$

Therefore, the conclusion holds for $n=k$.
By mathematical induction, the conclusion holds for $n \in \mathbf{N}_{+}$.
Since $2 \times 3^{6}<2011<3^{7}$, we have
$$
f(2011)=3 \times 2011-3^{7}=3846 .
$$