Saudi Arabian Mathematical Competitions

Taking $n = 1$, we obtain $\frac{1}{f(1)} = f(1)$. Because $f(1) > 0$, we deduce that $f(1) = 1$.

Taking $n = 2$, we obtain $f(3) + 1 = f(2)^2$, and taking $n = 3$ we obtain $4 f(3) + 4 + f(2) = f(2) f(3)^2$. Because $f(2) \neq 0$, this is equivalent to
$$f(3) + 1 = f(2)^2, \quad \text{ and } \quad 4 f(2) + 1 = f(3)^2$$
We deduce that $4 f(2) + 1 = (f(2)^2 - 1)^2$, and therefore
$$f(2)(f(2) - 2)(f(2)^2 + 2 f(2) + 2) = 0$$
But $f(2) \neq 0$. Hence $f(2) = 2$ and $f(3) = 3$.

Now assume, for all $1 \leq k \leq n$, that $f(k) = k$. We have
$$\frac{n-1}{n} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(n-1) \cdot n} = \frac{n}{f(n+1)} - \frac{1}{n \cdot f(n+1)} = \frac{n^2 - 1}{n f(n+1)}$$
and therefore $f(n+1) = n + 1$.

This proves that $f(n) = n$, for all $n \in \mathbb{N}$.