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For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3.

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Suppose that you add up the consecutive integers $n,n+1,\dots,n+m$; call the sum $S$. Now add up the numbers again, but in reverse numerical order, and arrange the sums like this:

$$\begin{array}{c} n&+&(n+1)&+&\ldots&+&(n+m-1)&+&(n+m)&=&S\\ (n+m)&+&(n+m-1)&+&\ldots&+&(n+1)&+&n&=&S\\ \hline (2n+m)&+&(2n+m)&+&\ldots&+&(2n+m)&+&(2n+m)&=&2S \end{array}$$

Each column must have the same sum: each time you move one column to the right, the top number increases by $1$ and the bottom number decreases by $1$, so the total remains the same. There are $m+1$ columns, so the bottom line can be summarized as

$$(m+1)(2n+m)=2S\;,$$

and $$S=\frac{(m+1)(2n+m)}2=\underbrace{(m+1)}_{\text{nr. of terms}}\cdot\frac{2n+m}2\;.$$

Since the number of terms is odd, $m+1$ is odd. But $S$ is an integer, so $(m+1)(2n+m)$ must be divisible by $2$. What does that tell you about $2n+m$?

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Formula for sum of consecutive integers: $S = \frac{1}{2}n(f + l)$ where $S$ is the sum, $n$ is the number of terms, $f$ is the first term and $l$ is the last.

$S$ is an integer. So, if $n$ is odd, then $\frac{1}{2}(f + l)$ has to be an integer [Reason: $\frac{1}{2}(f + l)$ can either be an integer or half integer, but half integer is not an option]. So $n$ divides $S$.

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Hint: Suppose there are $2t+1$ terms. Let $m$ be the "middle" term. Argue that the sum is $m$ times $2t+1$.

I suggest doing it by the "Gauss" method (which is older than Gauss by millenia) fpr summing an arithmetic progression. Go outwards from the middle. The two neighbours of the middle term add up to $2m$. So do the next two terms as you go outwards on both sides. And the next. And so on.

Remark: The same proof works for any arithmetic progression of integers with an odd number of terms.

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Hint $\rm\,\ m = 2n\!+\!1\:$ consecutive integers form a complete set of remainders mod $\rm\:m,\:$ so they are congruent mod $\rm\:m\:$ to the complete set of remainders $\rm\:0,\pm1,\pm2,\ldots,\pm n,\:$ which have sum $\equiv 0.$

Remark $\ $ This provides an arithmetical interpretation of the reflection around the middle term mentioned in some other answers. $ $ Here it becomes the $ $ negation $ $ reflection $\rm\ x \,\to\, -x \,\ (mod\ m),\:$ by which we can pair up each element with its negation, forcing them to cancel out of the sum. Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.

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