Saudi Arabian Mathematical Competitions

It follows $f(f(f(n)))=f(3 n)$, hence for all $n \in \mathbb{N}$ we have
$$f(3 n)=3 f(n) \tag{1}$$
For $n=0$ we find $f(0)=0$. We have $f(1) \neq 1$. Indeed, if $f(1)=1$, then we get $3=f(f(1))=f(1)=1$, not possible. Hence $f(1)>1$, so $3=f(f(1))>f(1)>1$. It follows $f(1)=2$, and consequently $f(2)=f(f(1))=3$.
Now we prove by induction that for all $n \in \mathbb{N}$, we have
$$f\left(3^{n}\right)=2 \cdot 3^{n} \text{ and } f\left(2 \cdot 3^{n}\right)=3^{n+1} \tag{2}$$
Indeed, the relations (2) hold for $n=0$. If they hold for a positive integer $n$, then we get
$$f\left(3^{n+1}\right)=f\left(3 \cdot 3^{n}\right)=3 f\left(3^{n}\right)=2 \cdot 3^{n+1}$$
and
$$f\left(2 \cdot 3^{n+1}\right)=3 f\left(2 \cdot 3^{n}\right)=3^{n+1}$$
There are exactly $3^{n}-1$ positive integers $k$ such that $3^{n}<k< 2 \cdot 3^{n}$. Also, there are exactly $3^{n}-1$ positive integers $k'$ such that
$$f\left(3^{n}\right)=2 \cdot 3^{n}<k'<3^{n+1}=f\left(2 \cdot 3^{n}\right)$$
The function $f$ is strictly increasing, then we have
$$f\left(3^{n}+k\right)=2 \cdot 3^{n}+k, \quad 0 \leq k \leq 3^{n}$$
hence $f\left(2 \cdot 3^{n}+k\right)=f\left(f\left(3^{n}+k\right)\right)=3\left(3^{n}+k\right)$.
In our case, we have $2010=3 \cdot 670$ and it follows
$$f(2010)=3 f(670)=3 f\left(2 \cdot 3^{5}+184\right)=9\left(3^{5}+184\right)=3843 .$$

Let $n=\overline{a_{1} a_{2} \ldots a_{t}}$ the representation of $n$ in base 3. We have, if $a_{1}=1$, then $f(n)=\overline{2 a_{2} \ldots a_{t}}$. If $a_{1}=2$, then $f(n)= \overline{1 a_{2} a_{3} \ldots a_{t} 0}$. These relations are direct consequences from (2). As in the previous solution we have $f(2010)=3 f(670)$. But $670=220211$, hence
$$\begin{gathered} f(670)=f\left(220211_{3}\right)=1202110_{3} \\ =3^{6}+2 \cdot 3^{5}+2 \cdot 3^{3}+3^{2}+3=1281, \end{gathered}$$
and we get $f(2010)=3 \cdot 1281=3843$.