This is a little novelty in magic squares. 
These squares may be formed with numbers that are in arithmetical progression, or that are not in such progression. 
If a square be formed of the former class, one place may be left vacant, but only under particular conditions. 
In the case of our puzzle there would be no difficulty in making the magic square with 9 missing; but with 1 missing (that is, using 2, 3, 4, 5, 6, 7, 8, and 9) it is not possible. 
But a glance at the original illustration will show that the numbers we have to deal with are not actually those just mentioned. 
The clown that has a 9 on his body is portrayed just at the moment when two balls which he is juggling are in mid-air. 
The positions of these balls clearly convert his figure into the recurring decimal .̍9. 
Now, since the recurring decimal .̍9 is equal to 9/9, and therefore to 1, it is evident that, although the clown who bears the figure 1 is absent, the man who bears the figure 9 by this simple artifice has for the occasion given his figure the value of the number 1. 
The troupe can consequently be grouped in the following manner:

Every column, every row, and each of the two diagonals now add up to 12. 
This is the correct solution to the puzzle.