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I’ve tried this.

Let the integers be $a_1, a_2,...,a_n$.

On dividing any number by $n$, we have $n$ possible remainders which are $0,1,2,...n-1$. Let these $n$ remainders be $n$ pigeonholes.

From the $n$ integers available, we can form $2^n - 1$ possible subsets, assuming each subset must contain at least one element. Each subset is assumed to be a pigeon.

$2^n - 1$ is always greater than or equal to $n$, for all values of $n$ greater than or equal to 1. So, the pigeonhole of the remainder 0, must contain at least one element.

That element is divisible by $n$. Now, when it contains more than one element, the sum of such elements will also be divisible by n, since the individual numbers are divisible by n.

I don’t think this is the right approach though. Any corrections and other methods will be appreciated.

Edit: This approach is completely wrong, as is my previous understanding of the pigeonhole principle. Answers and a link to a similar question are in the comments.

user733666
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    $2^n - 1$ is always greater than or equal to $n$, for all values of $n$ greater than or equal to 1. So, the pigeonhole of the remainder 0, must contain at least one element. >>> this is wrong, it shows you don't understand the pigeonhole principle – peter.petrov Sep 29 '20 at 16:08
  • :( Okay. I will study it again. – user733666 Sep 29 '20 at 16:09
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    "So the pigeonhole of the remainder $0$ must contain at least one element" Umm, no I don't think you can conclude that immediately from what you've written. What you can conclude however is that there must be at least two subsets who have the same remainder. Now... what can you say about their difference? Why doesn't that work right away? Can we choose to instead work with a very special class of subsets such that we can work with their differences? – JMoravitz Sep 29 '20 at 16:10
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    Recall, if we have $p$ pigeons and $h$ holes with $p>h$ and we place every pigeon into a hole, this implies that there must be at least one hole containing at least two pigeons. This doesn't say anything about which pigeons those happened to be in the crowded hole, nor does it say anything about which hole it was that was crowded. It doesn't say anything about whether or not there are even any pigeons at all in the "first" hole. All it says is that there is at least one hole (whichever hole it happened to be) with at least two pigeons (whatever pigeons those happened to be) in it. – JMoravitz Sep 29 '20 at 16:13
  • @JoséCarlosSantos it does!! Thank you. – user733666 Sep 29 '20 at 16:22
  • @JMoravitz Okay, so any of the pigeonholes (0,1,2,...,n-1) has at least two pigeons (subsets). We don’t know which pigeonhole. When two numbers leave the same remainder on division by $n$, their difference is divisible by $n$. So this would work for a problem that asked about the differences, and not the sum. Correct? – user733666 Sep 29 '20 at 16:27
  • "so this would work for a problem that asked about the differences, and not the sum" It just needs to be cleverly worded to work for our case and for us to keep track of what it is a difference of and what it is a sum of. Here, yes, it is a difference... but it is a difference of sums or a difference of sets depending on how you describe it... and what results is itself a sum. For instance $(5+9+11+18+32) - (5+9+11) = (18+32)$ – JMoravitz Sep 29 '20 at 16:56

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This part is incorrect:

$2^n-1$ is always greater than or equal to $n$, for all values of $n$ greater than or equal to $1$. So, the pigeonhole of the remainder $0$, must contain at least one element.

That's not what the pigeonhole principle states:

In mathematics, the pigeonhole principle states that if $n$ items are put into $m$ containers, with $n > m$, then at least one container must contain more than one item.

Solution:

Lets examine the following subsets: $\{a_1\}, \{a_1, a_2\}, \{a_1, a_2, a_3\}, ..., \{a_1, a_2, a_3\, ... , a_N\}$ and their sums' remainders. If any remainder is $0$ then we've solved the problem. Otherwise we have $N-1$ different remainders $(1, 2, 3, ..., N-1)$ and $N$ different subsets. By the pigeonhole principle we know that at least $2$ of those have the same remainder. Let those $2$ be $\{a_1, a_2, a_3\, ... , a_i\}$ and $\{a_1, a_2, a_3\, ... , a_j\}$ where $i < j$ and their remainder is $r$. We know that the sum of the subset $\{a_1, a_2, a_3\, ... , a_j\}$ is $\{a_{i+1}, , ... , a_j\}$ = $\{a_1, a_2, a_3\, ... , a_j\} - \{a_1, a_2, a_3\, ... , a_i\}$. Therefore the remainder is $r-r=0$ and it is divisible by $N$.

JMoravitz
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Daniel Atanasov
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