Solution:
Throughout this solution, assume we're working modulo $101$.

First, $N=1$. Let $\omega$ be a primitive $101$st root of unity. We then let $\omega_{n}=\omega^{1 / n}$, which we can do because $101$ is prime, so $1 / n$ exists for all nonzero $n$ and $1 / n=1 / m \Longrightarrow m=n$. Thus the set contains only one distinct element, $\omega$.

$M=100$ is impossible. Fix $\zeta$, a primitive $101$st root of unity, and let $\omega_{n}=\zeta^{\pi(n)}$ for each $n$. Suppose that there are $100$ distinct such $n \pi(n)$ exponents; then $\pi$ permutes the set $\{1,2, \cdots, 100\}$. Fix $g$, a primitive root of $101$; write $n=g^{e_{n}}$ and $\pi(n)=g^{\tau\left(e_{n}\right)}$. Then $\left\{e_{n}\right\}=\{0,1,2, \ldots, 100\}$ and $\tau$ is a permutation of this set, as is $e_{n}+\tau\left(e_{n}\right)$. However, this is impossible: $\sum_{n=1}^{100} e_{n}+\tau\left(e_{n}\right)=5050+5050 \equiv 5050\pmod{100}$, which is a contradiction. Thus there cannot be $100$ distinct exponents.

$M=99$ is possible. Again, let $\zeta$ be a primitive root of unity and let $\omega_{n}=\zeta^{1 /(n+1)}$, except when $n=100$, in which case let $\omega_{100}$ be the last possible root. Notice that $\frac{n}{n+1}=\frac{m}{m+1}$ if and only if $n=m$, so this will produce $99$ different elements in the set.

Thus $M-N=99-1=98$.