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 101st 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: $\sum_{n=1}^{100} e_{n}+\tau\left(e_{n}\right)=5050+5050 \equiv$ $5050(\bmod 100)$, 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 $M-N=99-1=98$.