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Incredibly, the result will always be \(n^2\).

To see why, imagine writing every number, \(n+1\leq k\leq 2n\), in the form $$k=2^ab$$ where \(b\) is an odd number and also the \(k\)'s largest odd factor. The next largest number whose largest odd factor is \(b\) will be \(2^{a+1}b=2k\). But this will be larger than \(2n\), so outside the range. Therefore each number in the range has a different largest odd factor.

Each of the largest odd factors must be one of \(1, 3, 5, ..., 2n-1\), as they cannot be larger than \(2n\). But there are \(n\) odd numbers here and \(n\) numbers in the range, so each number \(1, 3, 5, ..., 2n-1\) is the highest odd factor of one of the numbers (as the highest odd factors are all different).

Therefore, the sum of the odd factors is the sum of the first \(n\) odd numbers, which is \(n^2\).