Problem
ALG-B2-M01-P010 A closed set of numbers
There are \(2027\) real numbers written on a board. The sum of any three written numbers is also among the written numbers. Prove that at least \(2025\) of them are zero.
Hint 1. Order the numbers.
Hint 2. Compare the sum of the three smallest with the minimum and the sum of the three largest with the maximum.
Let \(a_1\le a_2\le\cdots\le a_{2027}\). Since \(a_1+a_2+a_3\) is written on the board, it is at least \(a_1\), hence \(a_2+a_3\ge0\). Similarly, \(a_{2025}+a_{2026}+a_{2027}\) is written and is at most \(a_{2027}\), hence \(a_{2025}+a_{2026}\le0\). But \(a_2\le a_3\le a_{2025}\le a_{2026}\), so \[0\le a_2+a_3\le a_{2025}+a_{2026}\le0.\] Thus \(a_2+a_3=a_{2025}+a_{2026}=0\), and the ordering gives \(a_2=a_3=\cdots=a_{2026}=0\). Hence there are at least \(2025\) zeros. The example \(-1,0,\ldots,0,1\) shows the bound is sharp.
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Regional level 7: one must use two extremal comparisons and squeeze the whole set between them.
G. Why this is not a one-step exercise. One choice of three numbers gives only half of the chain; both lower and upper bounds are needed.