First, for a pair of positive integers $(a, b)$ (where $a \neq b$), define one operation on it as replacing the larger number with the remainder obtained by dividing it by the smaller number. (For example $(2, 5)$ becomes $(2, 1)$, $(5, 2)$ becomes $(1, 2)$)
Let $g: Q^+ \to N_0$ be such that for every positive rational number $\frac{p}{q}$ (where $p, q$ are coprime positive integers), the following holds:
after performing the operation on the pair $(p, q)$ a total of $g(\frac{p}{q})$ times, one of the two numbers becomes 1 and the other is not 0.
First, since $p, q$ are coprime, $g$ exists for every positive rational number. Moreover, once one of them becomes 1, the next operation will make the other one become 0, and after that no further operation can be performed, so $g$ is unique. Hence $g$ is well-defined.
Next, it is not hard to see that $g(x) = g(x^{-1})$. And when $0 < x < 1$,
$$g(x+n) = g(x) + 1. \tag{1}$$
Here $n$ is a positive integer.

Next, we prove by mathematical induction that $f(\frac{p}{q}) = p$, where $p, q$ are coprime positive integers. We induct on $g(\frac{p}{q})$.

If $g(\frac{p}{q}) = 0$, then $p = 1$ or $q = 1$.
If $q = 1$, substituting $(x, n) = (1, p-1)$ into the original equation gives
$$f(p) = p. \tag{2}$$
If $p = 1$, substituting $(q, 1)$ into the original equation and combining with (2) gives $f(\frac{1}{q}) = 1$.
Thus the statement holds when $g(\frac{p}{q}) = 0$.

Suppose the statement holds when $g(\frac{p}{q}) = i - 1$; then when $g(\frac{p}{q}) = i$, we discuss two cases.

Case (1): $\frac{p}{q} > 1$. Let $n = \lfloor \frac{p}{q} \rfloor$, then by (1) we know $g(\frac{p-nq}{q}) = i-1$. Substituting $(\frac{p-nq}{q}, n)$ into the original equation and combining with (1) and the induction hypothesis, we get
$$f(\frac{p}{q}) = (p-nq) + nq = p. \tag{3}$$

Case (2): $\frac{p}{q} < 1$. Since $\frac{q}{p} > 1$, by (1) we know $g(\frac{p+q}{p}) = g(\frac{q}{p}) = i$.
Also $\frac{p+q}{p} > \frac{q}{p} > 1$, so by (3) we know $f(\frac{p+q}{p}) = p+q$, $f(\frac{q}{p}) = q$
Substituting $(\frac{q}{p}, 1)$ into the original equation gives $f(\frac{p}{q}) = (p+q) - q = p$.
Combining the above, the statement holds when $g(\frac{p}{q}) = i$. Hence by mathematical induction the statement always holds.
Therefore, $f(\frac{p}{q}) = p$, where $p, q$ are any coprime positive integers.