Judy sent us the following solution:

Suppose that we have a labelling with magic total T that looks like this:
A       E       I
B    D    F    H
    C        G
Then we can replace each entry x by 10-x and get another labelling (because if x is one of 1, 2, ..., 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 ł 6, so 4T ł 45+6=51, so T ł 13. Also, C+E+G £ 24, so 4T £ 45+24=69 so T £ 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=4×15-45=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.