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.

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.