Romanian Mathematical Olympiad

Obviously, $a_0 = -1$. For each $n \in \mathbb{N}^*$, denote by $k_n$ the number of terms among $a_1, a_2, \dots, a_n$ which are equal to $1$ (the other being $-1$). Then, the sequence $(x_n)_{n \ge 0}$ whose limit must be $\frac{1}{2}$ is $x_n = k_n\sqrt{n+1} - k_n\sqrt{n-1}$, or
$$x_n = \frac{2k_n}{\sqrt{n+1} + \sqrt{n-1}} = \frac{2\frac{k_n}{\sqrt{n}}}{\sqrt{1 + \frac{1}{n}} + \sqrt{1 - \frac{1}{n}}}.$$
It is enough to construct $(a_n)_{n \ge 1}$ such that $\lim_{n \to \infty} \frac{k_n}{\sqrt{n}} = \frac{1}{2}$. (*)
Let us choose $a_n = 1$ if and only if $n = (2m)^2$, with $m \in \mathbb{N}^*$. Then $(2k_n)^2 \le n < (2(k_n + 1))^2$, whence $\frac{\sqrt{n}}{2} - 1 < k_n \le \frac{\sqrt{n}}{2}$, which guarantees (*).