SAMC

It follows $f(f(f(n)))=f(3n)$, hence for all $n\in \mathbb{N}$ we have $$f(3n)=3f(n)\text{\hspace{1em}(1)}$$ For $n=0$ we find $f(0)=0$. We have $f(1)\ne 1$. 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 $$f\left(3^n\right)=2\cdot 3^n\text{ and }f\left(2\cdot 3^n\right)=3^{n+1}\text{\hspace{1em}(2)}$$ Indeed, the relations (2) hold for $n=0$. If they hold for a positive integer $n$, then we get $$f\left(3^{n+1}\right)=f\left(3\cdot 3^n\right)=3f\left(3^n\right)=2\cdot 3^{n+1}$$ and $$f\left(2\cdot 3^{n+1}\right)=3f\left(2\cdot 3^n\right)=3^{n+1}$$ 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 $k^{\prime }$ such that $$f\left(3^n\right)=2\cdot 3^n<k^{\prime }<3^{n+1}=f\left(2\cdot 3^n\right)$$ The function $f$ is strictly increasing, then we have $$f\left(3^n+k\right)=2\cdot 3^n+k,0\le k\le 3^n$$ 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 $$f(2010)=3f(670)=3f\left(2\cdot 3^5+184\right)=9\left(3^5+184\right)=3843.$$

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Alternative solution.

Let $n=\overline{a_1{a}_2\dots a_t}$ the representation of $n$ in base 3. We have, if $a_1=1$, then $f(n)=\overline{2a_2\dots a_t}$. If $a_1=2$, then $f(n)=\overline{1a_2{a}_3\dots a_{t}0}$. These relations are direct consequences from (2). As in the previous solution we have $f(2010)=3f(670)$. But $670=220211$, hence $$\begin{array}{c}f(670)=f\left(220211_3\right)=1202110_3\\ =3^6+2\cdot 3^5+2\cdot 3^3+3^2+3=1281,\end{array}$$ and we get $f(2010)=3\cdot 1281=3843$.