Answer:49^2 Let's renumber so that the initial arrangement is 1,...,99 Lower Bound: Let's take the sequence of numbers from 1 to n and reverse it. Let's prove by induction on n that for odd numbers we get ((n-1)/2)^2 here, and for even numbers we get (n/2)*(n/2-1). Let's take the original circle (there the numbers from 1 to n are counterclockwise), and in the final circle (let's see which of the n final circles we went to) the numbers from 1 to n are clockwise. Then it's not hard to prove that in the case of odd n there will be someone who has traveled a distance >=(n-1)/2, and in the case of even n at least n/2-1. That is, this number has made at least that many exchanges with the others. Now let's throw out this number and apply the induction assumption (after throwing it out, we are left with a permutation and an reversed permutation). Upper Bound: Let's prove that 49^2 is always possible. Let's move each object to the final position clockwise or counterclockwise. Then we have formed some arcs with orientations. Let's move each object at a constant speed from the start location to the end location in time 1. Then for every two elements there are the number of times they swapped places (0,1 or 2) which depend on the arcs. Let's spin the final circle, and choose for each element the minimum arc (size <=49) along which we will lead it at speed 1. Take any two different elements, and see how many times they swapped places when we spin the final circle. At no position of the final сircle could they have swapped places 2 times, because otherwise the sum of these arcs would be >99. Let the (minimum) distance between these two elements in the first circle be a, and in the second circle b. Then we will prove that they swapped at most in the 99-a-b positions of the circle. There are two cases when the orientations of the minimum arcs on the original circle and on the final circle coincide and vice versa. The first case is |a-b|<99-a-b (a,b<=49), the second case is exactly 99-a-b. Then for 99 positions of the final circle we have the sum of the number of exchanges on such strategy <=sum(i,j)(99-a-b)=99*C(99,2)-((49*50)/2)*99*2=99*99*99*49-49*50*99=49^2*99, that is, there is some with no more than 49^2 exchanges.

$2401=49^2.$

We prove turning from $1,2,\ldots ,99$ to $99,98,\ldots ,1$ needs at least $2401$ steps. This is because any triple $(i,j,k)$ will change there relative position iff two of them swaped. So if we construct a graph with $V=[99]$ and $i\sim j$ iff they didn't swap. So there are no triangle in this graph, by Turan's Theorem $|E|\le\lfloor 99^2/4\rfloor =49\times 50.$ Therefore we need at least $\binom{99}2-49\times 50=49^2$ moves.

The idea is really tricky. WLOG we say the final position is $1,2,\ldots ,99.$ Now call $1\sim 50$ small and $51\sim 99$ big. By Intermediate value theorem there exists $50$ consecutive number with $25$ big and $25$ small. The rest 49 numbers are 24 big and 25 small. Now we hope one of the two position appear: after this swap each area with 24(25) numbers with $24\times 49$ steps, and then swap numbers in small area and big area, we are done. The number of miminum step is actually exactly the number of reverse pair, they need $49\times 50$ steps in total, so one need at most $25\times 49$ steps, now $24\times 49+25\times 49=49^2.\Box$