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.

  • $\begingroup$ If you just pick them so that none of them are divisible by 2013, then certainly no subset of them of size 2013 will have any numbers divisible by 2013. $\endgroup$ – Brendan W. Sullivan Oct 1 '12 at 5:49
  • $\begingroup$ Did you mean "the sum of any 2013 of them is not divisible by 2013" ? $\endgroup$ – Brendan W. Sullivan Oct 1 '12 at 5:50
  • $\begingroup$ oh yeah sorry that's what i meant, now the typo is corrected $\endgroup$ – fmat Oct 2 '12 at 5:38

Half of them 1, half of them 2013.

  • $\begingroup$ I definitely think this is working well when we are allowed to choose an integer more than 1 time. But just wondering if the condition that all the 4024 numbers have to be distinct, is my way of thinking above still correct? thank you. $\endgroup$ – fmat Oct 2 '12 at 6:34
  • $\begingroup$ If you want them to be distinct, you can take the numbers $1+2013k$ and $2013k$ for $k=1,2,\dots,2012$. $\endgroup$ – Gerry Myerson Oct 2 '12 at 7:21

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