Solution:
We will show that $f(2001)$ must equal $3816$. We start by proving a lemma which gives us some of the values of $f(n)$.

Lemma: For $n = 0, 1, 2, \ldots$,
(a) $f\left(3^{n}\right) = 2 \cdot 3^{n}$; and
(b) $f\left(2 \cdot 3^{n}\right) = 3^{n+1}$.

Proof: We use induction. For $n = 0$, note that $f(1) \neq 1$, otherwise $3 = f(f(1)) = f(1) = 1$, which is impossible. Since $f(k)$ is a positive integer for all positive integers $k$, we conclude that $f(1) > 1$. Since $f(n+1) > f(n)$, $f$ is increasing. Thus $1 < f(1) < f(f(1)) = 3$ or $f(1) = 2$. Hence $f(2) = f(f(1)) = 3$.

Suppose that for some positive integer $n \geq 1$,
$$f\left(3^{n}\right) = 2 \cdot 3^{n} \text{ and } f\left(2 \cdot 3^{n}\right) = 3^{n+1}$$
Then,
$$f\left(3^{n+1}\right) = f\left(f\left(2 \cdot 3^{n}\right)\right) = 2 \cdot 3^{n+1}$$
and
$$f\left(2 \cdot 3^{n+1}\right) = f\left(f\left(3^{n+1}\right)\right) = 3^{n+2}$$
as desired. This completes the induction, and establishes the lemma.

Continuing with our solution, there are $3^{n} - 1$ integers $m$ such that $3^{n} < m < 2 \cdot 3^{n}$ and there are $3^{n} - 1$ integers $m'$ such that
$$f\left(3^{n}\right) = 2 \cdot 3^{n} < m' < 3^{n+1} = f\left(2 \cdot 3^{n}\right).$$
Since $f$ is an increasing function,
$$f\left(3^{n} + m\right) = 2 \cdot 3^{n} + m$$
for $0 \leq m \leq 3^{n}$. Therefore
$$f\left(2 \cdot 3^{n} + m\right) = f\left(f\left(3^{n} + m\right)\right) = 3\left(3^{n} + m\right)$$
for $0 \leq m \leq 3^{n}$. Hence
$$f(2001) = f\left(2 \cdot 3^{6} + 543\right) = 3\left(3^{6} + 543\right) = 3816.$$

Solution:
(Sketch) Andrew Dudzik's insight was to recognize that $f(n)$ deals in a very simple way, with the base-3 representation of $n$. Let $n = a_{1} a_{2} \cdots a_{t}$ be the base-3 digits of $n$. For example, if $n = 50$ in base 10, we would write $n = 1212$ in base 3 and hence $a_{1} = 1, a_{2} = 2, a_{3} = 1, a_{4} = 2$. Dudzik proved the following:

1. If $a_{1} = 1$, then $f(n) = 2 a_{2} a_{3} \cdots a_{t}$.
2. If $a_{1} = 2$, then $f(n) = 1 a_{2} a_{3} \cdots a_{t} 0$

These two statements can be proven easily with induction; we leave this as an exercise for the reader. Note that Dudzik's formulas allow us to immediately compute $f(2001)$. In base-3 notation, $2001$ is equal to $2202010$. Thus, by formula #2,
$$f(2202010) = 12020100 = 1 \cdot 3^{2} + 2 \cdot 3^{4} + 2 \cdot 3^{6} + 3^{7},$$
and this equals $3816$ in base-10 notation.