All-Soviet-Union Mathematical Olympiad

Solution:

Let the lightest container be $S$ weighing $s$. The other containers can be taken in 7 flights. The smallest load on these flights must be $\leq (18 - s) / 7$, and so that flight has spare capacity of at least $3 - (18 - s) / 7 = (3 + s) / 7$. Thus it can accommodate $S$ provided that $(3 + s) / 7 \geq s$, or $s \leq 1 / 2$.

At least one of the 7 flights takes $\leq 4$ containers. These weigh at most the weight of the 4 heaviest. Since the 31 lightest weigh at least $31s$, the 4 heaviest weigh at most $18 - 31s$. Thus this flight has spare capacity of at least $3 - (18 - 31s) = 31s - 15$ and can accommodate $S$ provided that $31s - 15 \geq s$, or $s \geq 1 / 2$.