Let $p$ and $q$ be the solutions the the equation $10x^2 - mx + 420 = 0$. We use the fact that the sum and product of the roots of a quadratic equation $ax^2+bx+c = 0$ are given by $-b/a$ and $c/a$, respectively, so $p+q = m/10$ and $pq = 420/10 = 42$. Since $m = 10(p+q)$, we minimize $m$ by minimizing the sum $p+q$. Since $p$ and $q$ are integers and multiply to 42, the possible values of $(p,q)$ are $(1,42),(2,21),(3,14),(6,7),(7,6),(14,3),(21,2),(42,1)$. (Note that if $p$ and $q$ are both negative, then $p+q$ is negative, so $m$ would be negative, which is excluded by the problem.) The sum $p+q$ is minimized when $(p,q) = (6,7)$ or $(7,6)$. In either case, $m = 10(p+q) = 10(6+7) =130
