All-Soviet-Union Mathematical Olympiad

Solution:

Put $x_{1} = x_{2} = \ldots = x_{10} = 3\pi /2$, $x_{11} = x_{12} = \ldots = x_{20} = \pi /2$. Then
$$sin x_{1} + \sin x_{2} + \ldots + \sin x_{20} = (-1 - 1 - 1 - \ldots - 1) + (1 + 1 + \ldots + 1) = 0,$$
and
$$sin x_{1} + 2 \sin x_{2} + \ldots + 20 \sin x_{20} = - (1 + 2 + \ldots + 10) + (11 + 12 + \ldots + 20) = 100.$$
So there is a solution with $n = 20$.

If there is a solution with $n < 20$, then there must be a solution for $n = 19$ (put any extra $x_{i} = 0$). But then
$$100 = (\sin x_{1} + 2 \sin x_{2} + \ldots + 19 \sin x_{19}) - 10 (\sin x_{1} + \sin x_{2} + \ldots + \sin x_{19})$$
$$= -9 \sin x_{1} - 8 \sin x_{2} - 7 \sin x_{3} - \ldots - \sin x_{9} + \sin x_{11} + 2 \sin x_{12} + \ldots + 9 \sin x_{19}.$$
But $|\text{rhs}| \leq (9 + 8 + \ldots + 1) + (1 + 2 + \ldots + 9) = 90$. Contradiction. So there is no solution for $n < 20$.