Romanian Mathematical Olympiad - Final Round

Without loss of generality, we may choose $a \le b \le c$.

The positive real numbers $a, b, c$ are such that $a + b + c = 1$, thus $a, b, c \in (0, 1)$. From $\frac{1}{a}, \frac{1}{b}, \frac{1}{c} \in (1, \infty)$ follows that $[\frac{1}{a}], [\frac{1}{b}], [\frac{1}{c}] \in \mathbb{N}^*$, therefore $n \ge 3$.

If $[\frac{1}{a}] \le 2$, then $a, b, c \in (\frac{1}{3}, 1)$ and $a + b + c > 1$, false. Therefore $[\frac{1}{a}] \ge 3$, so $n \ge 5$.
If 5 were an interesting number, we would have $[\frac{1}{a}] = 3$ and $[\frac{1}{b}] = [\frac{1}{c}] = 1$, whence $a+b+c > b+c > 1$, false.
If 6 were an interesting number, we would obtain $[\frac{1}{a}] = 3$, $[\frac{1}{b}] = 2$, $[\frac{1}{c}] = 1$, whence $a \in (\frac{1}{4}, \frac{1}{3}]$, $b \in (\frac{1}{3}, \frac{1}{2}]$ and $c \in (\frac{1}{2}, 1)$, therefore $a+b+c > \frac{1}{4}+\frac{1}{3}+\frac{1}{2} = \frac{13}{12}$, false.

We prove that all the natural numbers $n \ge 7$ are interesting.

Consider $k \in \mathbb{N}$, $k \ge 4$ and the real numbers $a = \frac{1}{k}, b = c = \frac{k-1}{2k}$, with $a+b+c=1$.
As $[\frac{1}{a}] = k, [\frac{1}{b}] = [\frac{1}{c}] = [2+\frac{2}{k-1}] = 2$, we have $[\frac{1}{a}] + [\frac{1}{b}] + [\frac{1}{c}] = k+4$. Consequently, all natural numbers $n \ge 8$ are interesting.
Choosing, for instance, $a = \frac{8}{30}, b = c = \frac{11}{30}$, we have $[\frac{1}{a}] + [\frac{1}{b}] + [\frac{1}{c}] = 3+2+2=7$ and $a+b+c=1$, therefore 7 is also an interesting number.

Thus, all natural numbers $n \ge 7$ are interesting.