Answer: $10! = 3628800$ ways.
Note that in a $10 \times 10$ grid, the number of ways to color $10$ cells red so that each row and each column has exactly one red cell is $10!$. Below we establish a bijection between such colorings of the $10 \times 10$ grid and calendars that comply with the regulations.

– Grid $\rightarrow$ calendar complying with the regulations:
As in the figure above, perform the following operations:

1. First look at two adjacent rows; if the red cell in the left row is higher than the red cell in the right row, then bold the grid line between the two rows.
2. Next look at two adjacent columns; if the red cell in the upper column is to the right of the red cell in the lower column, then bold the grid line between the two columns.
3. Erase all the bolded grid lines, and we obtain a calendar that complies with the regulations:

(Proof)
* First we prove that after erasing, the result is a calendar. Note that within each region separated out by the bold grid lines, there is at most one red cell; this is because if there were two, then if the relative positions of the two red cells are upper-left and lower-right, there must still be a bold vertical line between them, and if they are lower-left and upper-right, there must still be a bold horizontal line between them, and in either case this is a contradiction. This guarantees that after erasing the thin grid lines, the result must be a calendar.
* Next we prove that it complies with the regulations. Condition (1) is obviously satisfied. If condition (2) were violated, this would mean that in the original grid there must be a bold horizontal line such that the red cells in the row above it are all to the left of the red cells in the row below it, but this contradicts the very construction of the bold horizontal line itself. Condition (3) follows similarly.

### - Calendar complying with the regulations $\rightarrow$ grid:
For each cell in the calendar, perform the following actions in order:

1. Suppose there are $p$ red cells in the same row as this cell, and $q$ red cells in the same column, then subdivide this cell further into $q \times p$ smaller cells.
2. If this cell is red, and it is the $r$-th red cell counting from the top in its row, and the $r$-th small cell counting from the left in its column, then among the subdivided smaller cells, color red the cell in the $r$-th row counting from the bottom and the $s$-th column counting from the left, leaving the rest white.

It is easy to see that the above constitutes the inverse map of the mapping in the first part.