So, to start with, let’s

do it a technique and see what sum we get. Let the $a$s be (so as) $1,2,ldots,n$ and the $b$s be (so as) $2n,2n-1,ldots,n+1$. Then the variations are, so as, $2n-1,2n-3,ldots,1$ whose sum is $n^2$.

In view of

the title, presumably that is at all times the reply.

But why? I supply first a somewhat routine, prosaic resolution, the form of factor somebody who’s executed this form of factor earlier than is aware of they are going to have the ability to get to work. (This is the primary resolution I discovered.) Then one thing slicker. (This is the second resolution I discovered.) Then a approach to streamline the second a part of the slicker argument. And then a approach to streamline the primary half, discovered not by me however by person crazy walt in feedback. I want to depart all these items in, somewhat than simply presenting the slickest model up to now discovered, as a result of that is a extra sincere illustration of how issues usually work in arithmetic; see additionally this associated 3blue1brown video about calculation versus slick perception.

Here is a prosaic resolution; I think there’s something higher (and am excited about that). We can flip any legitimate partition into some other by repeatedly performing operations of the shape “discover two consecutive numbers one in all which is an $a$ and the opposite a $b$, and swap them”. What occurs once we do that? Suppose $a_i$ and $b_j$ are consecutive. Then not one of the different order relations change once we swap them, so our new sequences so as are the identical because the outdated besides that $a_i’=b_j$ and $b_j’=a_i$. If $i=j$ then the expression we’re thinking about does not change in any respect. Otherwise, say that $i<j$. Then $b_j simeq a_i<a_j$ the place $simeq$ means “differ by 1”, and subsequently $a_j>b_j$; and $a_isimeq b_j<b_i$, and subsequently $a_i<b_i$. So rising $a_i$ by 1 and reducing $b_j$ by 1 reduces $|a_i-b_i|$ by 1 and will increase $a_j-b_j$ by 1, and vice versa if as an alternative we lower $a_i$ and enhance $b_j$. In both case our expression does not change.

OK, now let’s examine if we are able to do one thing less complicated.

We have all these intervals from $a_i$ to $b_i$. Every quantity $ok$ from $1$ to $2n$ is an endpoint of precisely one in all these intervals; I declare that the primary $n$ are all left endpoints and the final $n$ are all proper endpoints. Why? Well, color all of the $a$s amber and all of the $b$s blue; suppose $ok$ is amber; if $ok=a_i$ then there are $i-1$ amber numbers to its left, therefore $k-i$ blue numbers to its left, therefore $n-k+i$ blue numbers to its proper, the primary of which is $b_{n-k+i}$. So $b_i$ is true of $a_i$ iff $ileq n-k+i$, iff $kleq n$. Likewise if $ok$ is blue.

And now

our sum is simply the sum over all $ok$ of the variety of these intervals it is in (counting 1/2 when it is precisely at an endpoint), which relies upon solely on the variety of left and proper endpoints on both aspect of $ok$, and we have simply seen that this by no means modifications.

Or (a basically-equivalent substitute for the foregoing paragraph):

our sum is simply the sum of all the proper endpoints minus the sum of all of the left endpoints, which is to say $bigl[(n+1)+cdots+2nbigr]-bigl[1+cdots+nbigr]=n^2$.

This is certainly slicker than the primary proof above, however I’d be unsurprised to search out that one can streamline issues additional.

In feedback, crazy walt suggests one other approach to do the primary bit:

i.e., proving that $1,…,n$ are left endpoints and $n+1,…,2n$ are proper endpoints. Consider the interval whose endpoints are $a_i,b_i$. We’ll show that its right-hand endpoint is $>n$. Note that $a_i$ is larger than $i-1$ smaller $a$s, and $b_i$ is larger than $n-i$ smaller $b$s. So whichever of them is bigger is larger than the opposite one (1), and $i-1$ smaller $a$s, and $n-i$ smaller $b$s, all of these units being disjoint; therefore larger than at the least $1+i-1+n-i=n$ different numbers; therefore it is $>n$.

(To my thoughts that is neater than my argument however extra rabbit-out-of-hat-ish.)