A square with horizontal and vertical sides is drawn on the plane. It held several segments parallel to the sides, and there are no two segments which lie on one line or intersect at an interior point for both segments. It turned out that the segments cuts square into rectangles, and any vertical line intersecting the square and not containing segments of the partition intersects exactly $ k $ rectangles of the partition, and any horizontal line intersecting the square and not containing segments of the partition intersects exactly $\ell$ rectangles. How much the number of rectangles can be?

[i]I. Bogdanov, D. Fon-Der-Flaass[/i]
1. **Restate the Problem:**
   Consider a square \( U = [0,1]^2 \) in the Cartesian coordinate system. The square is partitioned into rectangles by several segments parallel to the sides of the square. The conditions are:
   - Any vertical line \( x = a \) intersects exactly \( k \) rectangles.
   - Any horizontal line \( y = b \) intersects exactly \( \ell \) rectangles.
   - No two segments intersect at an interior point, and each line \( x = a \) or \( y = b \) contains at most one segment.

   We need to determine the maximum number of rectangles that can be formed under these conditions.

2. **Construct a Configuration with \( k\ell \) Rectangles:**
   - Consider \( k \) horizontal strips, each divided into \( \ell \) rectangles. This configuration results in \( k \times \ell = k\ell \) rectangles.
   - This construction satisfies the conditions since each vertical line intersects \( k \) rectangles and each horizontal line intersects \( \ell \) rectangles.

3. **Prove that \( k\ell \) is the Maximum Number of Rectangles:**
   - **Lemma:** Every segment \( s \) in the configuration has the same number \( a(s) \) of rectangles on either side of it.
     - **Proof:** Without loss of generality, assume \( s \) is a vertical segment. Let \( \mathcal{L} \) and \( \mathcal{R} \) be the sets of rectangles to the left and right of \( s \), respectively. Let \( \mathcal{X} \) be the set of rectangles that the vertical line containing \( s \) intersects. Consider two vertical lines \(\varepsilon\) distance away from \( s \), one to the left and one to the right. These lines intersect the rectangles in \( \mathcal{L} \) and \( \mathcal{R} \), and both intersect the rectangles in \( \mathcal{X} \). From the given conditions, \( |\mathcal{L}| = |\mathcal{R}| = k - |\mathcal{X}| =: a(s) \). \(\blacksquare\)

4. **Algorithm to Transform Any Configuration into the Trivial Configuration:**
   - Let \( N \) be the number of vertical segments \( s \) with \( a(s) > 1 \). The algorithm will decrease \( N \) in each step, eventually reaching a configuration where \( N = 0 \).
   - If \( N = 0 \), the configuration is the trivial one with \( k \) horizontal strips, each divided into \( \ell \) rectangles.
   - Consider the vertical segment \( s \) closest to the left edge \( e = \{0\} \times [0,1] \) with \( a(s) > 1 \). Let \( \mathcal{L} \) and \( \mathcal{R} \) be as defined earlier. Starting from any rectangle \( T_0 \) in \( \mathcal{L} \), trace a chain of rectangles forming a 'tube' \( \mathcal{T} \) until reaching \( e \). Form \( a \) such tubes \( \mathcal{T}_1, \mathcal{T}_2, \ldots, \mathcal{T}_a \).
   - Let \( R_1, R_2, \ldots, R_a \) be the rectangles in \( \mathcal{R} \) from top to bottom. Rearrange the tubes \( \mathcal{T}_i \) such that \( e_i \) continues in the edge between \( R_i \) and \( R_{i+1} \). Erase \( s \) and replace it with \( a \) new vertical edges to separate \( \mathcal{T}_i \) from \( R_i \).
   - This process maintains the conditions and decreases \( N \) by 1.

5. **Conclusion:**
   - The algorithm terminates when \( N = 0 \), resulting in the trivial configuration with \( k\ell \) rectangles.
   - Therefore, the maximum number of rectangles is \( k\ell \).

\(\blacksquare\)

The final answer is \( \boxed{ k\ell } \).