The first person must take either stool 9 or 17
(because of symmetry, it doesn't matter which). Assume they pick seat
9. The next person will pick seat 25, since it is the furthest from
seat 9. The next two people will take Seats one and 17. The next three
will occupy 5, 13, and 21. The next six will occupy 3, 7, 11, 15, 19,
and 23. This seats the maximum of 13 people, and no one is sitting next
to another person. If a seat other than 9 or 17 is chosen first, the
total bar patrons will be less than 13.
Suppose that a second pair of hands turns together with the wrong pair of hands, but then in the correct way. When the wrong pair is in the same position as the correct pair, this means that the time is shown in the right way. First look at the hour-hands that are at the twelve. One turns with the correct speed, the other with a speed that is twelve times as small. These two hands are in the same position again when the 'slow' hand has progressed x minutes. The fast hand then has progressed 60+x minutes. For the time x that passed, then holds: (60+x)/12 = x. This means that x = 5 5/11 minutes. For the minutes-hands that start at six holds the same. The confused clock therefore shows the correct time again at 5 5/11 minutes past 7.
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