The answer is $24$. If we look at a 3x3 square first, it is easy to see that all six row sums can be odd only if 3, 5 or all 9 of the labels are odd, and that otherwise, there is at least one even row sum in each of the two directions. So in a cube, the 18 row sums in three parallel square layers (without considering the 9 rows perpendicular to them) can only be all odd if the number of odd labels in each layer is among 3, 5, 9. But we have 14 odd labels, and 14 cannot be partitioned into three of $\{3,5,9\}$. So there is at least one even row sum in each of the three (!) directions, which restricts the total of odd sums to at most 24. This maximum can be achieved, for example: ooo oxx oxx ooo xxx xox ooo xxo xxo Here "o" means an odd number. Only the three rows containing the center of the cube have even sums.