jmc

In the quadratic $a{x}^2+bx+c$, the roots sum to $\frac{-b}{a}$ and multiply to $\frac{c}{a}$. Therefore, for $x^2-nx+m$, we know that the sum of the roots is $n$ and the product of the roots is $m$. The requirement that $n$ be an integer with $0<n<100$ along with the requirement that the roots are consecutive positive integers leaves us with 49 possible values of $n$: $(1+2),(2+3),(3+4),...,(48+49),(49+50)$. Of these values of $n$, the corresponding value of $m$ would be $(1\ast 2),(2\ast 3),(3\ast 4),...,(48\ast 49),(49\ast 50)$. In order for $m$ to be divisible by 3, thus, one of the roots has to be divisible by three. The values of $n$ in $(2+3),(3+4),(5+6),...,(48+49)$ satisfy that requirement, but in $(1+2),(4+5),...,(49+50)$ they do not. This eliminates one third of the possibilities for $n$. Since $n=(49+50)$ is disqualified, we are left with $48-(48÷3)=48-16=\boxed{32}$ possible values of $n$.