jmc

If $n\ge 4$, then $$n^2-3n+2=(n-1)(n-2)$$is the product of two integers greater than 1, and thus is not prime. For $n=1$, $2$, and $3$ we have, respectively, $$(1-1)(1-2)=0,(2-1)(2-2)=0,\text{and}(3-1)(3-2)=2.$$Therefore, $n^2-3n+2$ is prime only when $n=3$, for a total of $\boxed{1}$ positive integer $n$.