Estimate. We will show that the number of integers never exceeds $50$.
We will track the differences between each number and the next one in clockwise order. If three consecutive numbers were $a$, $b$, and $c$, their differences were $a-b$ and $b-c$. After applying the operation to $b$, the numbers become $a$, $a-b+c$, and $c$, with differences $a - (a-b+c) = b-c$ and $(a-b+c)-c = a-b$. Thus, the operation simply swaps two adjacent differences. Initially, all differences were non-integer, so they remain non-integer at all times. Therefore, two integers can never appear consecutively, meaning their count cannot exceed $50$.

Example. First, arrange the numbers $0.1$ and $0.2$ alternately in a circle. If we perform the operation on each $0.2$, it will be replaced by $0.1 - 0.2 + 0.1 = 0$, making every other number an integer.

Now, we need to adjust the example so that all numbers become distinct. To do this, it is sufficient to add a small unique value to each $0.1$ and to each $0.2$ the sum of the values added to its neighbors. For instance, choosing $t = 0.001$, we can add to consecutive $0.1$s the numbers $0, t, 2t, \dots, 47t, 48t, 50t$; then, to the $0.2$s, we add $t, 3t, 5t, \dots, 95t, 98t, 50t$. As a result, all numbers will become distinct.

The explicitly constructed example looks as follows:
$$0.1 \ 0.2001 \ 0.1001 \ 0.2003 \ 0.1002 \ 0.2005 \ 0.1003 \ \dots \\ \dots \ 0.1047 \ 0.2095 \ 0.1048 \ 0.2098 \ 0.105 \ 0.205$$