Brazilian Math Olympiad

The answer is all real numbers in the interval $]1, \lfloor n/2 \rfloor[$. For simplicity, let
$$E = \frac{x_1}{x_n + x_1 + x_2} + \frac{x_2}{x_1 + x_2 + x_3} + \frac{x_3}{x_2 + x_3 + x_4} + \dots + \frac{x_n}{x_{n-1} + x_n + x_1}.$$
Let's prove first the lower bound. Let $S = x_1 + x_2 + \dots + x_n$. First notice that
$$E > \frac{x_1}{S} + \frac{x_2}{S} + \frac{x_3}{S} + \dots + \frac{x_n}{S} = \frac{x_1 + x_2 + \dots + x_n}{S} = 1.$$
By making, say, $x_i = \epsilon^{i-1}$, we obtain $\frac{x_i}{x_{i-1}+x_i+x_{i+1}} = \frac{\epsilon}{1+\epsilon+\epsilon^2}$ for $1 < i < n$, $\frac{x_1}{x_n+x_1+x_2} = \frac{1}{\epsilon^{n-1}+1+\epsilon}$ and $\frac{x_n}{x_{n-1}+x_n+x_1} = \frac{\epsilon^{n-1}}{\epsilon^{n-2}+\epsilon^{n-1}+1}$. Thus, by making $\epsilon$ very small we obtain $E$ arbitrarily close to 1.

Now, for the upper bound, notice that, for $n$ even,
$$E < \frac{x_1}{x_1 + x_2} + \frac{x_2}{x_1 + x_2} + \frac{x_3}{x_3 + x_4} + \frac{x_4}{x_3 + x_4} + \dots + \frac{x_n}{x_{n-1} + x_n} = \frac{n}{2} = \lfloor \frac{n}{2} \rfloor$$
For $n$ odd, suppose without loss of generality that the minimum of the denominators $x_n + x_1 + x_2, x_1 + x_2 + x_3, \dots, x_{n-1} + x_n + x_1$ is $x_1 + x_2 + x_3$. Thus
$$\frac{x_1}{x_n + x_1 + x_2} + \frac{x_2}{x_1 + x_2 + x_3} + \frac{x_3}{x_2 + x_3 + x_4} \le \frac{x_1}{x_1 + x_2 + x_3} + \frac{x_2}{x_1 + x_2 + x_3} + \frac{x_3}{x_1 + x_2 + x_3} = 1$$
This implies
$$\begin{aligned} E < 1 &+ \frac{x_4}{x_4+x_5} + \frac{x_5}{x_4+x_5} + \frac{x_6}{x_6+x_7} + \frac{x_7}{x_6+x_7} + \dots + \frac{x_{n-1}}{x_{n-1}+x_n} + \frac{x_n}{x_{n-1}+x_n} \\ = 1 &+ \frac{n-3}{2} = \lfloor \frac{n}{2} \rfloor \end{aligned}$$
Now, to attain the upper bound, choose $x_{2k} = \epsilon$ and $x_{2k-1} = 1$, $k = 1, 2, \dots, \lfloor n/2 \rfloor$. For $n$ even, we have the summands $\frac{x_{2k-1}}{x_{2k-2}+x_{2k-1}+x_{2k}} = \frac{1}{1+2\epsilon}$ and $\frac{x_{2k}}{x_{2k-1}+x_{2k}+x_{2k+1}} = \frac{\epsilon}{2+\epsilon}$ and $E$ gets arbitrarily close to $n/2$ as $\epsilon$ gets small. For $n$ odd, notice that $x_1 = x_n = \epsilon$, so $\frac{x_{2k-1}}{x_{2k-2}+x_{2k-1}+x_{2k}} = \frac{1}{1+2\epsilon}$ and $\frac{x_{2k}}{x_{2k-1}+x_{2k}+x_{2k+1}} = \frac{\epsilon}{2+\epsilon}$ and $E$ gets arbitrarily close to $\lfloor n/2 \rfloor$ as $\epsilon$ gets small.