jmc

Let $r$ be the common root, so $$\begin{array}{rl}{r}^2-ar+24& =0,\\ r^2-br+36& =0.\end{array}$$Subtracting these equations, we get $(a-b)r+12=0,$ so $r=\frac{12}{b-a}.$ Substituting into $x^2-ax+24=0,$ we get $$\frac{144}{(b-a{)}^2}-a\cdot \frac{12}{b-a}+24=0.$$Then $$144-12a(b-a)+24(b-a{)}^2=0,$$so $12-a(b-a)+2(b-a{)}^2=0.$ Then $$a(b-a)-2(b-a{)}^2=12,$$which factors as $(b-a)(3a-2b)=12.$

Let $n=b-a,$ which must be a factor of 12. Then $3a-2b=\frac{12}{n}.$ Solving for $a$ and $b,$ we find $$a=2n+\frac{12}{n},b=3n+\frac{12}{n}.$$Since $n$ is a factor of 12, $\frac{12}{n}$ is also an integer, which means $a$ and $b$ are integers.

Thus, we can take $n$ as any of the 12 divisors of 12 (including positive and negative divisors), leading to $\boxed{12}$ pairs $(a,b).$