Cut $2003$ disjoint rectangles from an acute-angled triangle $ABC$, such that any of them has a parallel side to  $AB$ and the sum of their areas is maximal.
1. **Understanding the Problem:**
   We need to cut 2003 disjoint rectangles from an acute-angled triangle \(ABC\) such that each rectangle has a side parallel to \(AB\) and the sum of their areas is maximal.

2. **Configuration of Rectangles:**
   The rectangles must be stacked one on top of the other, with the first rectangle resting on side \(AB\), the next rectangle resting on the first, and so on. Let the heights of these rectangles be \(x_1, x_2, \ldots, x_n\), where \(n = 2003\).

3. **Sum of Heights:**
   The sum of the heights of the rectangles must be less than or equal to the height \(h\) of the triangle:
   \[
   0 \leq \sum_{k=1}^n x_k = x \leq h
   \]

4. **Area Calculation:**
   The area of each rectangle is proportional to its height and the length of the side parallel to \(AB\). The sum of the areas of the rectangles can be expressed as:
   \[
   \text{Total Area} = \frac{a}{h} \left( h \sum_{k=1}^n x_k - \sum_{k=1}^n x_k^2 - \sum_{1 \leq i < j \leq n} x_i x_j \right)
   \]
   Simplifying this expression, we get:
   \[
   \text{Total Area} = \frac{a}{2h} \left( 2h \sum_{k=1}^n x_k - \sum_{k=1}^n x_k^2 - \left( \sum_{k=1}^n x_k \right)^2 \right)
   \]

5. **Optimization:**
   To maximize the total area, we use the method of Lagrange multipliers. The maximum of this expression is found to be:
   \[
   \frac{a}{2nh} x (2nh - (n+1)x)
   \]
   This occurs when \(x_1 = x_2 = \cdots = x_n = \frac{x}{n}\).

6. **Maximizing \(x\):**
   To maximize the expression, we set:
   \[
   x = \frac{n}{n+1} h
   \]
   Substituting this value into the expression for the total area, we get:
   \[
   \text{Total Area} = \frac{n}{n+1} \cdot \frac{ah}{2} = \frac{n}{n+1} \cdot \text{area}(\triangle ABC)
   \]

The final answer is \( \boxed{ \frac{n}{n+1} \cdot \text{area}(\triangle ABC) } \)