Among the points corresponding to number $1,2,...,2n$ on the real line, $n$ are colored in blue and $n$ in red. Let $a_1,a_2,...,a_n$ be the blue points and $b_1,b_2,...,b_n$ be the red points. Prove that the sum $\mid a_1-b_1\mid+...+\mid a_n-b_n\mid$ does not depend on coloring , and compute its value. :roll:
1. **Restate the problem with additional details:**
   We are given $2n$ points on the real line, numbered $1, 2, \ldots, 2n$. Among these, $n$ points are colored blue and $n$ points are colored red. Let $a_1, a_2, \ldots, a_n$ be the blue points in increasing order, and $b_1, b_2, \ldots, b_n$ be the red points in decreasing order. We need to prove that the sum $\mid a_1 - b_1 \mid + \mid a_2 - b_2 \mid + \ldots + \mid a_n - b_n \mid$ does not depend on the specific coloring and compute its value.

2. **Pairing and ordering:**
   Since $a_1, a_2, \ldots, a_n$ are in increasing order and $b_1, b_2, \ldots, b_n$ are in decreasing order, each pair $(a_i, b_i)$ will consist of one number from the set $\{1, 2, \ldots, n\}$ and one number from the set $\{n+1, n+2, \ldots, 2n\}$.

3. **Sum of absolute differences:**
   Consider the sum of absolute differences:
   \[
   \sum_{i=1}^n \mid a_i - b_i \mid
   \]
   Since $a_i \le n$ and $b_i \ge n+1$ (or vice versa), we can write:
   \[
   \mid a_i - b_i \mid = b_i - a_i
   \]
   because $b_i > a_i$ for all $i$.

4. **Sum of the differences:**
   We need to compute:
   \[
   \sum_{i=1}^n (b_i - a_i)
   \]
   Since $a_1, a_2, \ldots, a_n$ are the numbers $1, 2, \ldots, n$ in some order, and $b_1, b_2, \ldots, b_n$ are the numbers $n+1, n+2, \ldots, 2n$ in some order, we have:
   \[
   \sum_{i=1}^n a_i = 1 + 2 + \ldots + n = \frac{n(n+1)}{2}
   \]
   and
   \[
   \sum_{i=1}^n b_i = (n+1) + (n+2) + \ldots + 2n = \frac{n(3n+1)}{2}
   \]

5. **Compute the sum of differences:**
   Therefore,
   \[
   \sum_{i=1}^n (b_i - a_i) = \sum_{i=1}^n b_i - \sum_{i=1}^n a_i = \frac{n(3n+1)}{2} - \frac{n(n+1)}{2}
   \]
   Simplifying this, we get:
   \[
   \frac{n(3n+1) - n(n+1)}{2} = \frac{n(3n+1 - n-1)}{2} = \frac{n(2n)}{2} = n^2
   \]

6. **Conclusion:**
   Thus, the sum $\mid a_1 - b_1 \mid + \mid a_2 - b_2 \mid + \ldots + \mid a_n - b_n \mid$ is always $n^2$ regardless of the specific coloring.

\[
\boxed{n^2}
\]