Let $f(n)$ be a function satisfying the following three conditions for all positive integers $n$:

1. $f(n)$ is a positive integer.
2. $f(n + 1) > f(n)$.
3. $f(f(n)) = 3n$.

Find $f(2001)$.

Let's think step by step and output the final answer within \boxed{}.
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