Prove that the value of the expression $|a_1-b_1|+|a_2-b_2|+...+|a_n-b_n|$ does not depend on the coloring. Now we have some $n$ of the number from the set $\{1,2,...,2n\}$ colored red and the rest of them are colored blue. Say $a_1<a_2<...<a_n$ are red and $b_1>b_2>...>b_n$ are blue. Prove that the value of the expression $$E=|a_1-b_1|+|a_2-b_2|+...+|a_n-b_n|$$ does not depend on the coloring. 

All I can do is to calculate this $E$ if we take $a_i=i$ and $b_i = 2n+1-i$ for all $i\leq n$. In this case we get $$E = (n+1)+(n+2)+...+(2n) -1-2-...-n = n+n+...n = n^2$$ 
Clearly it wants us to prove that $E$ is invariant for such colorings. Does any one has any idea how to prove it. Strong assumption is that it should be done with induction.  
 A: Firstly, note that $|a-b|=\max\{a,b\}-\min\{a,b\}$ for all real $a$ and $b$. Hence,
$$
E=\sum_{k=1}^{n}\max\{a_k,b_k\}-\sum_{k=1}^{n}\min\{a_k,b_k\}.
$$
The main idea is to prove that actually $\max\{a_1,b_1\},\max\{a_2,b_2\},\ldots,\max\{a_n,b_n\}$ is a permutation of $n+1,n+2,\ldots,2n$ and similarly $\min\{a_1,b_1\},\min\{a_2,b_2\},\ldots,\min\{a_n,b_n\}$ is a permutation of $1,2,\ldots,n$ (clearly, it's enough to prove the first statement, the second follows immediately).
Since numbers $\max\{a_k,b_k\}$ are different elements of $\{1,2,\ldots,2n\}$ it's sufficient to prove that for all $k$, $1\leq k\leq n$ we have $\max\{a_k,b_k\}>n$.
Indeed, otherwise $a_k\leq n$ and $b_k\leq n$, so due to $a_1<a_2<\ldots<a_n$ and $b_1>b_2>\ldots>b_n$ we have $n+1$ numbers: $a_1,a_2,\ldots,a_k$ and $b_k,b_{k+1},\ldots,b_n$ which are not greater than $n$. However, it's impossible because  $a_1,\ldots,a_n,b_1,\ldots,b_n$ is a permutation of $1,2,\ldots,2n$ and among $1,2,\ldots,2n$ there are only $n$ numbers which are not greater than $n$. Thus, for all $k\in\{1,2,\ldots,n\}$ we have $\max\{a_k,b_k\}>n$, so $\max\{a_1,b_1\},\max\{a_2,b_2\},\ldots,\max\{a_n,b_n\}$ is a permutation of $n+1,n+2,\ldots,2n$, as desired.
Therefore,
$$
E=\sum_{k=1}^{n}\max\{a_k,b_k\}-\sum_{k=1}^{n}\min\{a_k,b_k\}=
\\
=((n+1)+(n+2)+\ldots+2n)-(1+2+\ldots+n)=n^2.
$$
Hence, $E=n^2$, as stated.
A: Ahh, this is what I found right now. For every $k$ we have $$a_{k+1}-a_k >0 >b_{k+1}-b_k$$
so $$ a_{k+1}-b_{k+1}> a_k-b_k$$
This should kill the problem...?
A: This is not a complete proof but I think something like this should work!
Let's prove a more general statement for any $2n$ consecutive integers $E$ doesn't depend on coloring/partitioning. Proceed by induction on $n.$ The bases cases are trivial to check. Assume that the assert holds for $\leq n-1$ no matter if we choose all $a_i$s first or $b_i$s first.
Now note that, we have either $a_1=1$ or $b_n=1.$ Also either $b_1=2n$ or $a_n=2n$
Case 1. $a_1=1$ and $b_1=2n,$ then $E$ is $(2n-1) + \sum (...)$ so by hypothesis $E$ doesn't depend on the partition.
Case 2. $a_1=1$ and $a_n=2n,$ then $E$ is $(b_1 - 1 ) + \sum (...) + (2n - b_n)
 = (b_1 - b_n) + (2n-1) + \sum (...).$ By hypothesis, it only remains to show that $b_1 - b_n$ doesn't depend on coloring as well. We can also assume that we have selected $a_i$s first, then $b_i$ were determined uniquely if that help!
Other two cases are similar as well. 
