Can you leave out one of the numbers $1, 2, \ldots, 2008$ in such a way that the remaining 2007 numbers have an arrangement $a_{1}, a_{2}, \ldots, a_{2007}$ where the differences $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2006}-a_{2007}\right|,\left|a_{2007}-a_{1}\right|$ are all distinct?
Solution. Since the number of differences in question is 2008, and we have 2007 numbers left, it seems reasonable to first write down all 2008 numbers alternately - for example, starting with the largest, then the smallest, and so on. Then we check which difference might appear more than once, and accordingly try to omit one number.

$2008,1,2007,2,2006, \ldots, 1508,501,1507,502,1506, \ldots, 1006,1003,1005,1004$.

The sequence of differences:

$$
2007,2006,, 2005, \ldots, 1007,1006,1005,1004, \ldots, 3,2,1,1004
$$

Then the number 1004 appears twice as a difference. This is, for example, the difference between 502 and 1506. Let's omit 502. Then the sequence is:

$2008,1,2007,2,2006, \ldots, 1508,501,1507,503,1506, \ldots, 1003,1006,1004,1005$.

And the differences are:

$$
2007,2006,, 2005,2004, \ldots, 1007,1006,1004,1003, \ldots, 3,2,1,1003
$$

It can be seen that 1003 appears twice, so this is not good.

Omitting 1506, the sequence is:

$$
2008,1,2007,2,2006, \ldots, 1507,502,1505,503,1504, \ldots, 1005,1003,1004
$$

And the sequence of differences is:

$$
2007,2006,, 2005,2004, \ldots, 1005,1003,1002,1001, \ldots, 2,1,1004
$$

This is therefore a valid solution.

Remarks. 1. Many solvers omitted 503 from the same sequence, which is also a valid solution.

2. If we start the sequence with 1 instead of 2008, we can similarly obtain that by omitting 503, we arrive at a valid sequence. Omitting 1506 also results in a valid solution in this case.