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RU-ALLRUSSIAN 1988 Problem 20

    Let m, n, k be positive integers with m≥n\mathrm{m} \geq \mathrm{n} and 1+2+…+n=mk1 + 2 + \ldots + \mathrm{n} = \mathrm{mk} . Prove that the numbers 1, 2, ..., n can be divided into k groups in such a way that the sum of the numbers in each group equals m.

    Dataset text NuminaMath olympiads. Quoted with source.

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