Is it only 1?
Resulting in:
 = \frac{n}{2} (n-1) )
Now to prove that this is the only trip possible to yield maximal length:
Our initial moves are
This is the only way we can get a length of '(n-1)', if we make a different move first, say 1 -> 2, and then
is NOT correct, since we already have been at 1.
Therefore the initial moves must be
The next move can be either to 2, 3, 4, .... , n-1
Likewise, we need to make a move of length (n-2) in order for the sum to work. The only way to do this is going to floor number 2, if we try to make the (n-1) length later on in the chain, we arrive at thet same scenario as the first. Therefore the next move must be:
Likewise, we proceed inductively, so that the only way to yield a move (n-3) in length, is to proceed from 2 -> n-1.
And so on, until we yeild a series of lengths:
 + (n-2) + \dots + 1 = \frac{n}{2} (n-1) )
We cannot permute the sum as proven above.
Now consider a general sum of moves to get to
 )
 )
There are n terms in the series because we need to make (n-1) moves to get to the other (n-1) floors.
) )
The only series that can therefore add up to
is
)
And since we have shown that we cannot permute this sum, therefore the only permutation allowed is:
 )
Therefore we only have 1 trip of maximal length.
Likewise 1 trip of minimal length with a similar method, 1 + 1 + \dots + 1
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