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Choose n numbers from the set \[\{1;2;...;2n\}\] and sort them in that way: \[a_1<a_2<...<a_n\] Sort the other n numbers so that \[b_1>b_2>...>b_n\]
Prove \[|a_1-b_1|+|a_2-b_2|+...+|a_n-b_n|=n^2\]

We claim:
$\sum_{i=1}^n |a_i-b_i|=2n+(2n-1)+....+(n+1)-n-(n-1)-(n-2)....-1=n^2$
Let $|a_i-b_i|=a_i-b_i$ for some $i$. Then $a_i>b_i>...>b_n$ and also $a_i>a_{i-1}>a_{i-2}>....>a_1$ it implies $a_i$ is greater than $(n-i+1)+(i-1)=n$ different elements from the set. It implies $a_i>n$. Thus we can prove that for $|a_i-b_i|=b_i-a_i$ that $b_i>n$ as then $b_i>a_i>a_{i-1}>...>a_1$ and $b_i>b_{i+1}>...>b_n$ same argument as before. It implies $max(a_i,b_i)>n$ hence $\sum_{i=1}^n |a_i-b_i|=(\sum_{i=n+1}^{2n} i)-(\sum_{j=1}^n j)=n^2$

done.
Cool problem Zan. Wecome to BDMO forum.

Thank you for your reply. But could you explain the last conclusion, please?
My teacher is very demanding. I've got many other unsolved problems.

sourav das wrote:It implies $max(a_i,b_i)>n$ hence $\sum_{i=1}^n |a_i-b_i|=(\sum_{i=n+1}^{2n} i)-(\sum_{j=1}^n j)=n^2$

For each $(a_i,b_i)$ exactly one of them is greater than $n$, and thus we can say that, all of the numbers greater than $n$ has "$+$" sign, and all the numbers from $1$ to $n$ has "$-$" sign and thus the sum $\sum_{i=1}^n |a_i-b_i|$ has the value $(\sum_{i=n+1}^{2n} i)-(\sum_{j=1}^n j)=n^2$

Thanks a lot!!!!