To this end, write $\ell = f(1)$, and notice that, since $1 \le f(n)/n \le \ell$, there exists a minimal positive integer $k \le \ell$ such that $\lfloor f(n)/n \rfloor = k$ for infinitely many positive integers $n$. Let $A$ denote the set of all these positive integers, let $B = \{n: n \in A \text{ and } 2n \notin A\}$, and let $A' = A \setminus B$.

We now show that $B$ is finite, so $A'$ is infinite. Indeed, $f(2n) \le 2f(n)$ and $\lfloor f(2n)/(2n) \rfloor \le \lfloor 2f(n)/(2n) \rfloor = k$ imply $\lfloor f(2n)/(2n) \rfloor < k$ for all $n$ in $B$, so $B$ is finite by minimality of $k$.

Next, for $n$ in $A'$, write
$$\frac{2f(n)}{f(2n)} = \frac{\frac{f(n)}{n}}{\frac{f(2n)}{2n}} < 1 + \frac{1}{k},$$
to deduce that the positive integer $2f(n)/f(2n)$ is less than $2$, so $f(2n) = 2f(n)$.

Further, fix a positive integer $a$ and notice, as before by minimality of $k$, that $f(a+n) \ge k(a+n)$ for all but finitely many $n$ in $A'$. Hence
$$\frac{f(a) + f(n)}{f(a + n)} < \frac{f(a) + (k + 1)n}{k(a + n)},$$
for all but finitely many $n$ in $A'$, so $(f(a) + f(n))/f(a+n)$ is a positive integer less than $2$ for all but finitely many $n$ in $A'$, i.e., $f(a+n) = f(a)+f(n)$ for all but finitely many $n$ in $A'$.

Finally, fix two positive integers $a$ and $b$. By the preceding, $f(a+n) = f(a)+f(n)$, $f(b+n) = f(b)+f(n)$, and $(f(a+n)+f(b+n))/f(a+b+2n)$ is a positive integer for all but finitely many $n$ in $A'$. Consequently,
$$\frac{f(a) + f(b) + 2f(n)}{f(a + b) + f(2n)} = \frac{f(a) + f(b) + f(2n)}{f(a + b) + f(2n)}$$
is a positive integer for all but finitely many $n$ in $A'$, so the latter must equal $1$ for all but finitely many $n$ in $A'$, i.e., $f(a+b) = f(a)+f(b)$. This ends the proof.