There exists a subsequence of every $n$ consecutive natural numbers whose sum is divisible by $n(n+1)/2$ This is my first question here, hence I apologize beforehand for any mistakes I make.
The statement is very simple.

For any natural $n$, show that for all sequences of $n$ consecutive natural numbers, there exists a subsequence with non-zero length whose sum is divisible by $(1+2+.. +n)$.

I really can't find any way to proceed with this question. I thought of double induction since this looks kind of analogous to $n$ consecutive natural numbers being divisible by $n!$. But I couldn't proceed anywhere with that. It looks like a question meant to be solved by utilization of pigeonhole principle, but I can't make my pigeonholes either. Any hint on how to proceed would be very appreciated.
 A: Note: I'm assuming "subsequence" doesn't require contiguity, as otherwise the problem statement is false for $n=3$ and $(2,3,4)$. So, I'm interpreting "subsequence" to be essentially the same as "subset."
We first treat the case where $n$ is odd. In this case, the numbers form a complete residue system modulo $n$, and we can group them into $\frac{n+1}2$ groups, each of which has sum a multiple of $n$ -- one of the groups contains only the number which is a multiple of $n$, and the other $\frac{n-1}2$ of which consist of pairs, pairing the number $\equiv i\pmod n$ with the number $\equiv -i\pmod n$. Let these groups be $S_1,\dots,S_{(n+1)/2}$, and consider the $\frac{n+1}2+1$ subsets
$$\emptyset\subset S_1\subset S_1\cup S_2\subset \cdots\subset S_1\cup S_2\cup\cdots\cup S_{\frac{n+1}2}.$$
These subsets each have sum divisible by $n$, and so there must be two which are equivalent modulo $\frac{n(n+1)}2$ by the pigeonhole principle. The difference between these two sets, which is $S_i\cup \cdots\cup S_j$ for some $i<j$, has sum divisible by $\frac{n+1}2$.
In the case that $n$ is even, the numbers form a complete residue system modulo $n+1$ with one exception. If that exception is $0$ modulo $n+1$, group the numbers into $n/2$ pairs, pairing $i$ with $-i$ modulo $n+1$. If not, put the number which is $0$ modulo $n+1$ in its own group, find $n/2-1$ remaining $(i,-i)$ pairs, and neglect the last unpaired element. Either way, we now have $n/2$ disjoint subsets $S_1,\dots,S_{n/2}$ of the $n$ consecutive numbers, each of which has sum a multiple of $n+1$, and we can finish as before.
