, 2023

Solution:

We perform casework by placing the entries from largest to smallest.

- The grid must have exactly one $5$ since an entry equal to $5$ will be the maximum in its row and in its column. We can place this in $9$ ways.
- An entry equal to $4$ must be in the same row or column as the $5$; otherwise, it will be recorded twice, so we only have two records left but $1,2$, and $3$ are all unrecorded. Using similar logic, there is at most one $4$ in the grid. So there are $4$ ways to place the $4$.
- We further split into cases for the $3$ entries. Without loss of generality, say the $4$ and the $5$ are in the same row.

- If there is a $3$ in the same row as the $4$ and the $5$, then it remains to label a $2 \times 3$ grid with $1$s and $2$s such that there is exactly one row with all $1$s, of which there are $2\left(2^{3}-1\right)=14$ ways to do so.

- Suppose there is no $3$ in the same row as the $4$ and the $5$. Then there are two remaining empty rows to place a $3$.

There are two possible places we could have a record of $2$, the remaining unoccupied row or the remaining unoccupied column. There are $2$ ways to pick one of these; without loss of generality, we pick the row. Then the column must be filled with all $1$s, and the remaining slots in the row with record $2$ can be filled in one of $3$ ways ($12$, $21$, or $22$). The final empty cell can be filled with a $1,2$, or $3$, for a total of $3$ ways. Our total here is $2 \cdot 2 \cdot 3 \cdot 5=60$ ways.

Hence, our final answer is $9 \cdot 4 \cdot (14+60)=36 \cdot 74=2664$.