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As stated on the title,

Is it possible to find $4024$ positive integers such that the sum of any $2013$ of them is not divisible by $2013$?

I used to assumed they have to be distinct. Being inspired by the answer below, I am thinking maybe we have to select those $4024$ numbers by two types: the first group of $2012$ integers being those that are divisible by $2013$, and the rest $2012$ integers are those not divisible by $2013$. So that if you pick any $2013$ of them you will end up having to pick at least $1$ of those $2012$ not-divisible-by-$2013$ integers. Does that work out?

If the numbers don't have to be distinct then the following suggestion seems to work well, but I am not sure whether they have to be distinct. I will update the question if there are any changes. Thank you guys.

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    oh yeah sorry that's what i meant, now the typo is corrected2012-10-02

1 Answers 1

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Half of them 1, half of them 2013.

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    If you want them to be distinct, you can take the numbers $1+2013k$ and $2013k$ for $k=1,2,\dots,2012$.2012-10-02