smc

Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right. Case 1: Center 4 $$\begin{array}{ccc}1& 2& \\ 3& 4& 8\\ & & 9\end{array}\begin{array}{ccc}1& 2& \\ 3& 4& \\ & 8& 9\end{array}$$ 3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2\ast 6=12$ Case 2: Center 5 $$\begin{array}{ccc}1& 2& 3\\ 4& 5& \\ & 8& 9\end{array}\begin{array}{ccc}1& 2& \\ 3& 5& \\ & 8& 9\end{array}\begin{array}{ccc}1& 2& \\ 3& 5& 8\\ & & 9\end{array}\begin{array}{ccc}1& 2& 3\\ 4& 5& 8\\ & & 9\end{array}$$ Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2\ast 9=18$ * Case 3: Center 6 By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2\ast 6=12$ $$12+18+12=\boxed{42}$$