Judy sent us the following solution:

Suppose that we have a labelling with magic total $T$ that looks like this:

$A\quad\quad E\quad\quad I$

$\ B\quad D\quad F\quad H$

$\quad C\quad \quad G\quad $

Then we can replace each entry $x$ by $10-x$ and get another labelling (because if $x$ is one of 1, 2, \ldots, 9, then $10-x$ will also be). Also, this label will have magic total $30-T$, because instead of adding up $A+B+C$, for example, we'll add up $10-A+10-B+10-C=30-T$, and all of the sums will be the same. So if we have a labelling with magic total $T$, then we certainly have one with magic total $30-T$.

Using the solution to the Magic W problem, we know that there is 1 magic labelling for $T=13$, and there are 5 for $T=14$. So (using what we worked out above) there's also one for $T=17$ and 5 for $T=16$. (There can't be any more for $T=17$, for example, because any labelling of $T=17$ is also one of $T=13$.)

When do magic labellings exist? Well, again using the ideas from the solution to the Magic W problem, we must have $C+E+G+45=4T$. But $C+E+G\geq 6$, so $4T\geq 45+6=51$, so $T\geq 13$. Also, $C+E+G\leq 24$, so $4T\leq 45+24=69$ so $T\leq 17$. So we only need to check whether there are any magic labellings for $T=15$. Suppose that there is one. Then we have $C+D+E=15$, and also $C+E+G=4\times 15-45=15$, so $D=G$. But that's not allowed, so there are no labellings of $T=15$.

To summarise, there are magic labellings for $T=13, 14, 16$ and $17$ (and no others), and there are $1+5+5+1=12$ magic labellings.