11. (20 points) $f(n)$ is defined on $N_{+}$, and is monotonically increasing, with its range also in $N_{+}$, and it satisfies $f(f(n))=3 n$. Find $f(2202)-$ $f(2022)$
Assume $f(1)=1$, then $f(f(1))=3=f(1)$, contradiction
Assume $f(1) \geq 3$, then by monotonicity $f(f(1)) \geq f(3)>3$, but $f(f(1))=3$, contradiction
(5 points)

Therefore, $1<f(1)<3$, so $f(1)=2$
Additionally, for $f(f(n))=3n$, taking $n=f(n)$, we have $3f(n)=f(f(f(n)))=f(3n)$, thus $f\left(3^{n}\right)=3^{n} \cdot f(1)=$
$2 \cdot 3^{n}$ $\qquad$
Taking $n=f\left(3^{n}\right)$, we have $f\left(2 \cdot 3^{n}\right)=f\left(f\left(3^{n}\right)\right)=3 \cdot 3^{n}$
Therefore, $f\left(2 \cdot 3^{n}\right)-f\left(3^{n}\right)=3^{n}$
Also, by monotonicity, $f\left(3^{n}+k\right)=2 \cdot 3^{n}+k\left(k \in N_{+}, k<3^{n}\right)$
(15 points)
$$
f\left(2 \cdot 3^{n}+k\right)=f\left(f\left(3^{n}+k\right)\right)=3\left(3^{n}+k\right)=3^{n+1}+3 k\left(k \in N_{+}, k<3^{n}\right)
$$

Thus, $f(2202)-f(2022)=f\left(3^{7}+15\right)-f\left(2 \cdot 3^{6}+564\right)=2 \cdot 3^{7}+15-3^{7}-564=1638$
(20 points)