[ Weighing ]

There are 2018 weights with masses of 1 g, 2 g, ..., 2018 g. The Hare placed two weights on one pan of the scales. The Wolf tried to balance them with two other weights on the other pan, but failed. Which weights could the Hare have taken?

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1) Explain why the Wolf will not be able to balance the specified sets. In the pairs of weights (1 g; 2 g) and (1 g; 3 g), the sum of the masses is less than the sum of the masses of any two weights from the remaining set, while in the pairs (2016 g; 2018 g) and (2017 g; 2018 g), on the contrary, the sum of the masses of the weights is greater than the sum of the masses of any two weights from the remaining set.
2) Show how the Wolf can balance any other pair of weights. Suppose the Hare chose weights with masses $m$ and $n$, where $n > m$. Then, if $n - m > 2$, the weights will be balanced by weights with masses $m + 1$ and $n - 1$ (such weights will definitely be found in the remaining set).

If $n - m = 1$ or $n - m = 2$, then they will be balanced by weights with masses $m - 1$ and $n + 1$. Such weights will be found in the remaining set if $m \neq 1$ and $n \neq 2018$, which are exactly the cases indicated in the answer.

## Answer

$(1$ g; 2 g); (1 g; 3 g); (2016 g; 2018 g); (2017 g; 2018 g $)$.