Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

We shall prove that the identity is the unique function satisfying the conditions in the statement. Clearly, $f(1) = 1$, so $f(n) \ge 2$ if $n \ge 2$, by injectivity. We will use the following well-known result.

Egyptian fractions theorem. For every positive rational $r$ and positive integer $n$, there exists a finite set $S$ of integers greater than $n$ such that $r = \sum_{s \in S} 1/s$.

Now, consider an integer $n \ge 2$ and use the Egyptian fractions theorem to write
$$1 - \frac{1}{n} = \sum_{s \in S} \frac{1}{s}, \text{ where } S \text{ is a finite set of integers greater than } n(n+1), \text{ and}$$
get thereby
$$1 = \frac{1}{n} + \sum_{s \in S} \frac{1}{s} = \frac{1}{n+1} + \frac{1}{n(n+1)} + \sum_{s \in S} \frac{1}{s}.$$
Both are positive integers, so
$$\frac{1}{f(n+1)} + \frac{1}{f(n(n+1))} - \frac{1}{f(n)}$$
is an integer. Since $-\frac{1}{2} \le -\frac{1}{f(n)} < \frac{1}{f(n+1)} + \frac{1}{f(n(n+1))} - \frac{1}{f(n)} < \frac{1}{f(n+1)} + \frac{1}{f(n(n+1))} \le \frac{1}{2} + \frac{1}{2} = 1$, it follows that $\frac{1}{f(n)} = \frac{1}{f(n+1)} + \frac{1}{f(n(n+1))}$. In particular, $f$ is strictly increasing, so $f(n) \ge n$.

Finally, proceed by induction on $n \ge 2$ to prove that $f(n) = n$. To show that $f(2) = 2$, simply notice that $2/f(2) = 1/f(2) + 1/f(3) + 1/f(6)$ is a positive integer not exceeding 1. To complete the proof, let $f(n) = n$ for some $n \ge 2$ and write
$$\frac{1}{n} = \frac{1}{f(n)} = \frac{1}{f(n+1)} + \frac{1}{f(n(n+1))} \le \frac{1}{n+1} + \frac{1}{n(n+1)} = \frac{1}{n}$$
to conclude that $f(n+1) = n+1$.