Really nice problem! Notice that the number $1$ is never in a good pair. By replacing 100 by smaller numbers $n$ we notice that the number of good pairs is $n-3$. We try to prove this for any $n$. We do this by a clever induction. Suppose that there are $n-3$ good pairs for the case of $n$ numbers on the circle, now I'll prove that the number of good pairs for $n+1$ numbers on a circle. Look at these numbers excluding the number 1. The pairs that were good for $n+1$ stay good and now we have $n-3$ good pairs without the one (by induction) but by adding back the 1, we get another good pair, i.e. the two neighbours of $1$ thus a total of $n-2$ good pairs and we're done. (Base case $n=3$ is easy to get $0$).