Consecutive Tallying

Relevant wiki: Sum of n, n², or n³

Let $S_n = \underbrace{n + (n+1)+(n+2) + ... + (2n-1)}_{n \text{ integers}}$ .

$\begin{aligned} \implies S_n & = \frac{n(\color{#3D99F6}{a+l})}{2} \quad \quad \small \color{#3D99F6}{\text{where }a \text{ and }l \text{ are the first and last integers of the series respectively.}} \\ & = \frac{n(n+2n-1)}{2} \\ & = \frac{n(3n-1)}{2} \\ \implies \frac{S_n}{n} & = \frac{3n-1}{2} \begin{cases} \color{#D61F06}{n = \text{even}} & \implies 3n - 1 \text{ is odd} & \color{#D61F06}{\implies n \not{|} \ S_n} \\ \color{#3D99F6}{n = \text{odd}} & \implies 3n - 1 \text{ is even} & \color{#3D99F6}{\implies n \ | \ S_n} \end{cases} \end{aligned}$

Therefore, $\boxed{\text{for every odd }n}$ , the sum of $n$ consequent positive integers is always divisible by $n$ .

Relevant wiki: Sum of n, n², or n³

The sum of $n$ consecutive numbers is only divisible by $n$ when $n$ is odd. Proof :
Sum of $n$ consecutive numbers $= a+(a+1)+(a+2)+(a+3) ... +(a+(n-1))=n⋅a+\sum _{ t=1 }^{ n-1 }{ t } =n⋅a+\frac { { (n-1) }^{ 2 }+(n-1) }{ 2 } =n⋅(a+\frac { n-1 }{ 2 } )$ it follows that the sum of $n$ consecutive numbers is always divisible by $n$ , when $n-1$ is even, so that the fraction $\frac { n-1 }{ 2 }$ becomes an integer. Therefore the sum of $n$ consecutive numbers is only divisible by $n$ when $n$ is odd. $_\square$

With $n$ odd the sum $S_n$ of the first $n$ elements is divisible by $n$ , because $n+1$ is even and $S_n = \frac{n(n+1)}{2} = nk$ , with $k$ integer, so "No values of $n$ satisfy this condition" is discarded. With $n = 2$ the sum of every two consecutive elements will be odd, then "for every $n$ " and "for every even $n$ " are discarded. Therefore the answer is "for every odd $n$ ".

I know this is not a very nice solution.