Let a sequence $n, n+1, ..., 2n-1$ of natural numbers be given, $n$ is odd. Prove that you can always remove a number from this sequence such that the sum of the remaining numbers is not divisible by any number that was originally in the sequence.
I found that the formula for the sum of the sequence is $S_n=\frac{1}{2}(3n-1)n$. I think if we remove the last number, so $S'_n=\frac{1}{2}(3n-1)n-2n+1 = \frac{1}{2}(3n-2)(n-1)$, it isn't divisible by any of the rest numbers. But I don't have any idea to prove that.
With $n=2k+1$, $\frac{1}{2}(3n-1)n=(3k+1)(2k+1)$. Now let us define two sets of $n$ consecutive numbers.
$a=\{(2k+1),(2k+2), ...,(4k+1)\}$
$b=\{\left((3k+1)(2k+1)-(4k+1)\right),\left((3k+1)(2k+1)-(4k)\right),...,\left((3k+1)(2k+1)-(2k+1)\right)\}$
Now here is the key. Every element of $a$ divide at most one element of $b$, but $(3k+1)(2k+1)-(2k+1)=(3k)(2k+1)$ is divisible by $(3k)$ and $(2k+1)$, both of which are element of $a$.
Consequently, there is at least one element of $b$ that is not divisible by any element of $a$.