Choose any N numbers from the set { 1, 2, 3, ..., 2N }, and write them in a row in increasing order.

Then write the remaining numbers in a second row below, in decreasing order.

Then, in a third row, write the absolute values of the differences between the corresponding numbers in row 1 and row2.

Finally, add up the numbers in the third row.

Write the first two rows as:

a_{1} |
a_{2} |
a_{3} |
... |
a_{N-1} |
a_{N} |

b_{1} |
b_{2} |
b_{3} |
... |
b_{N-1} |
b_{N} |

We claim that for each i, one of {a_{i}, b_{i}} is in { 1, 2, ..., N } and the
other is in { N+1, N+2, ..., 2N }.
If not, then we can find a pair {a_{i}, b_{i}} whose elements are both in { 1, 2, ..., N }
or both in { N+1, N+2, ..., 2N }.

If a_{i} and b_{i} are both in { 1, 2, ..., N }, then we have 1 <= a_{1} < a_{2} < ... < a_{i} <= N
and N >= b_{i} > b_{i+1} > ... > b_{N} >= 1.
But this gives N+1 distinct integers between 1 and N inclusive, which is impossible.
A similar argument shows that a_{i} and b_{i} cannot both be in { N+1, N+2, ..., 2N }.

Hence if we take the larger of the corresponding a and b values, we just get N+1, N+2, ..., 2N in some order,
taking the smaller of the corresponding a and b values gives 1, 2, ..., N some order.
The sum of the absolute values of the differences between the a and b values is thus (N+1) + (N+2) + ... + 2N - (1 + 2 + ... + N) = N + N + ... + N = N^{2}.

This result is due to Proizvolov.