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Old 09-12-2007, 03:50 PM   #2
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We have
$\sum\limit_{i = 1}^{2n}|a_{i + 1} - a_{i}| + \sum\limit_{i = 1}^{2n}|a_{i + 1} + a_i| = 2\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} $
Thus, $\sum\limit_{i = 1}^{2n}|a_{i + 1} - a_i| = 2\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} - 2\sum\limit_{i = 1}^{2n}a_i $. Hence, $(\sum\limit_{i = 1}^{2n - 1}|a_{i + 1} - a_i|) + |a_1 - a_{2n}| = 2\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} - 2n(2n + 1) $. (*)

Because $2n - 1 $ numbers $|a_{i + 1} - a_i| $ are distinct and $|a_{i + 1} - a_i|\le 2n - 1 $. Then,
$\sum\limit_{i = 1}^{2n - 1}|a_{i + 1} - a_i| = 1 + 2 + ... + 2n - 1 = n(2n - 1) $. (**)
From (*) and (**) we have $|a_1 - a_{2n}| = 2\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} - 2n(2n + 1) - n(2n - 1) $.
$= 2\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} - 4n^2 - n $.

If, $a_1 - a_{2n} = n $ we have $\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} = 3n^2 + n $. (1)

But, $\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\}\le 2(2n + 2n - 1 + ... + n + 1) = 2n(2n + 1) - n(n + 1) = 3n^2 + n $ (2)

From $(1); (2) $we have either $\{a_1;a_3;...;a_{2n - 1}\} = \{1;2;..;n\} $ or $\{a_2;...;a_{2n}\} = \{1;2;...;n\} $.

But $a_1 - a_{2n} = n $ hence, $\{a_2;...;a_{2n}\} = \{1;2;...;n\} $

If $\{a_2;..;a_{2n}\} = \{1;2;...;n\} $ so, $\sum\limit_{i = 1}^{2n}max\{a_{i + 1};a_i\} = 3n^2 + n $.
Hence, $|a_1 - a_{n}| = 2(3n^2 + n) - 4n^2 - n = n $ therefore, $a_1 - a_{2n} = n $

It looks easy
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