Pick a spot for $4$. Then it must be the greatest in its row, so we must have two of 1, 2, 3 in the row with it. Pick which is one spot to its left, and which is one spot to its right. (If the rows wrap around, of course.) Then the last of 1, 2, 3 must not be the same column as 4, or else 4 isn't the smallest in its column. So pick one of the four squares remaining for it. Then automatically $a=4$ and $b=4$, so we can just place the other 5 numbers anywhere. Using this scheme, with all choices are independent, we find that the total number is $9\cdot3\cdot2\cdot4\cdot5!=25920$.