On an $n$ × $n$ board, the set of all squares that are located on or below the main diagonal of the board is called the$n-ladder$. For example, the following figure shows a $3-ladder$:
[asy]
draw((0,0)--(0,3));
draw((0,0)--(3,0));
draw((0,1)--(3,1));
draw((1,0)--(1,3));
draw((0,2)--(2,2));
draw((2,0)--(2,2));
draw((0,3)--(1,3));
draw((3,0)--(3,1));
[/asy]
In how many ways can a $99-ladder$ be divided into some rectangles, which have their sides on grid lines, in such a way that all the rectangles have distinct areas?
1. **Understanding the Problem:**
   We need to divide a $99$-ladder into rectangles such that all rectangles have distinct areas. The $99$-ladder consists of all squares on or below the main diagonal of a $99 \times 99$ board.

2. **Counting the Total Number of Squares:**
   The total number of squares in an $n$-ladder is given by the sum of the first $n$ natural numbers:
   \[
   1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
   \]
   For $n = 99$, the total number of squares is:
   \[
   \frac{99 \times 100}{2} = 4950
   \]

3. **Distinct Areas Requirement:**
   Since all rectangles must have distinct areas, the areas must be the first $99$ natural numbers: $1, 2, 3, \ldots, 99$.

4. **Placement of Rectangles:**
   Each rectangle must have a unique area, and the only way to ensure this is to place them in a specific manner. The largest rectangle (area $99$) must be placed first, and it can only be placed in two ways: either horizontally or vertically.

5. **Recursive Placement:**
   After placing the largest rectangle, we are left with a $(99-1)$-ladder, i.e., a $98$-ladder. The same logic applies recursively to the $98$-ladder, $97$-ladder, and so on.

6. **Counting the Number of Ways:**
   For each step, there are $2$ choices (horizontal or vertical placement). Therefore, the total number of ways to place all rectangles is:
   \[
   2^{n-1}
   \]
   For $n = 99$, the number of ways is:
   \[
   2^{99-1} = 2^{98}
   \]

Conclusion:
\[
\boxed{2^{98}}
\]