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  "body": "\n## Statement\n\nLet $n$ be a positive integer.\nThen among any $2n - 1$ integers, not necessarily distinct, there are $n$ whose sum is divisible by $n$.\n\n## Remarks\n\nThe proof first treats a prime $n = p$, using the Chevalley–Warning theorem or the [[Cauchy–Davenport theorem|F7K7XFACSG]].\nFor composite $n = ab$, the cases $a$ and $b$ imply the case $n$.\nOne repeatedly extracts $2b - 1$ disjoint groups of $a$ elements, each with sum divisible by $a$.\nThen one applies the case $b$ to the $2b - 1$ group sums divided by $a$.\nThe number $2n - 1$ cannot be lowered.\nThe $2n - 2$ integers consisting of $n - 1$ zeros and $n - 1$ ones have no $n$ elements with sum divisible by $n$.\nIndeed, any $n$ of them contain $j$ ones with $1 \\le j \\le n - 1$, so their sum $j$ is not a multiple of $n$.\nFor $n = 2$: among any three integers, two have the same parity, so their sum is even.\n\n## Explanations\n\n### L1 human\n\nWrite down $2n - 1$ whole numbers.\nYou can always choose exactly $n$ of them whose sum is a multiple of $n$.\nFor example, among any $5$ integers, some $3$ add up to a multiple of $3$.\n\n### L2 human\n\nThe theorem is the starting point of zero-sum theory in [[additive combinatorics|3NM362A2JG]].\nIn the language of the [[integers modulo $n$|03VFZ2WETT]], every sequence of $2n - 1$ elements of $\\Z/n\\Z$ has a subsequence of length $n$ with sum $0$.\nRelated invariants, such as the Davenport constant, ask for the shortest length that forces a nonempty zero-sum subsequence.\n",
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