Original Problem
I came across this problem here. (Section 3 - Question 3):
An der Tafel stehen 11 natürliche Zahlen. Zeige, dass man aus diesen Zahlen einige (vielleicht alle) wählen und dazwischen die Zeichen + und − so platzieren kann, dass das Ergebnis durch 2011 teilbar ist.
I translate this as:
Given 11 natural numbers, show that one can choose some numbers (maybe all) such that by placing + and - between them, you can get a number divisible by 2011.
Provided Solution(?)
I found the answer here, but it does not explain anything to me. (See my confusion below)
Wir betrachten alle mögliche Summen. Jede dieser 11 Zahlen kann zur Summe gehören oder nicht, also gibt es $2048 = 2^{11}$ Summen. Nach dem Schubfachprinzip werden irgendwelche zwei Summen S1 und S2 den gleichen Rest bei der Division durch 2011 ergeben. Dann ist die Differenz von S1 und S2 durch 2011 teilbar. Nach Kürzen der Zahlen, die in beiden Summen vorhanden sind, ist entweder S1 − S2 oder S2 − S1 von der gewünschten Form.
Translation:
We consider all possible sums. Each of these 11 numbers may or may not belong to the sum, so there are $2048 = 2^{11}$ sums. According to the pigeonhole principle, any two sums S1 and S2 will give the same remainder when divided by 2011. Then the difference of S1 and S2 is divisible by 2011. After shortening the numbers present in both sums, either $S1 - S2$ or $S2 - S1$ is of the desired form.
Question
Now, I am confused because we are not given any constraints for the natural numbers. But then one can choose the numbers $\{1,2,3 \dots 11\}$, where the sum, 66 is les then 2011, so clearly not divisible by it either. Moreover, where does the above solution come from? Is this just a mistake in my translation?
Note: Ofcourse, you could just choose natural numbers, which would always satisfy the need (multiples of 2011, for example), but I believe that violates the spirit of the question's real intent.
The numbers could sum to $0$ i.e., in your example $\{1,2,3,\ldots, 11\}$, take $3-2-1=0$ which is divisible by $2011$.
This should read: According to the Pigeonhole Principle, there are [as opposed to "any"] two sums $S_1$ and $S_2$ such that $S_1 \pmod{2011} = S_2 \pmod{2011}$. Then $S_1-S_2$ can be written in the desired form. But then these things happen in translations from one language to another.