NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Let us consider $x_1, x_2, \dots, x_n$ a solution of the equation above. We may suppose, without loss of generality, that $x_1 + x_2 + \dots + x_n \ge 0$ (otherwise we change the signs of all the numbers) and that $x_1 \le \dots \le x_p \le 0 < x_{p+1} \le \dots \le x_n$. Then the equation becomes $-2(x_1+x_2+\dots+x_p) = m$. From $x_i \in (-1, 1)$ we obtain that $m < 2p$, i.e. $p > \frac{m}{2}$. Moreover, $n-p > x_{p+1} + \dots + x_n \ge -x_1 - \dots - x_p = \frac{m}{2}$. If $m$ is even then $p \ge \frac{m}{2}+1$, $n-p \ge \frac{m}{2}+1$, and therefore $n \ge m+2$. A convenient choice for $n = m+2$ is $x_1 = \dots = x_{\frac{m}{2}+1} = -\frac{m}{m+2}$, $x_{\frac{m}{2}+2} = \dots = x_n = \frac{m}{m+2}$.

If $m$ is odd, from $p \ge \frac{m+1}{2}$, $n-p \ge \frac{m+1}{2}$, we obtain $n \ge m+1$. A convenient choice for $n = m+1$ is $x_1 = \dots = x_{\frac{m+1}{2}} = -\frac{m}{m+1}$, $x_{\frac{m+1}{2}+1} = \dots = x_n = \frac{m}{m+1}$.

In conclusion, the smallest value of $n$ for which the equation has solutions is
$n = m + 1$ if $m$ is odd and $n = m + 2$ if $m$ is even.