$56$ lines are drawn on a plane such that no three of them are concurrent. If the lines intersect at exactly $594$ points, what is the maximum number of them that could have the same slope? 
1. **Define the problem and variables:**
   We are given 56 lines on a plane with no three lines being concurrent, and they intersect at exactly 594 points. We need to find the maximum number of lines that could have the same slope (i.e., be parallel).

2. **Set up the equation:**
   Suppose \( x \) of the lines are parallel. These \( x \) lines do not intersect each other. The remaining \( 56 - x \) lines intersect each other at most \( \binom{56-x}{2} \) times. Additionally, each of these \( 56 - x \) lines intersects each of the \( x \) parallel lines exactly once.

3. **Calculate the number of intersections:**
   - The number of intersections among the \( 56 - x \) lines is:
     \[
     \binom{56-x}{2} = \frac{(56-x)(55-x)}{2}
     \]
   - The number of intersections between the \( 56 - x \) lines and the \( x \) parallel lines is:
     \[
     (56 - x)x
     \]

4. **Set up the total number of intersections:**
   The total number of intersections is given by:
   \[
   \frac{(56-x)(55-x)}{2} + (56 - x)x = 594
   \]

5. **Simplify the equation:**
   Multiply both sides by 2 to clear the fraction:
   \[
   (56-x)(55-x) + 2(56 - x)x = 1188
   \]
   Expand and simplify:
   \[
   3080 - 111x + x^2 + 112x - 2x^2 = 1188
   \]
   Combine like terms:
   \[
   3080 + x - x^2 = 1188
   \]
   Rearrange to form a quadratic equation:
   \[
   -x^2 + x + 1892 = 0
   \]

6. **Solve the quadratic equation:**
   Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{-1 \pm \sqrt{1^2 - 4(-1)(1892)}}{2(-1)}
   \]
   Simplify under the square root:
   \[
   x = \frac{-1 \pm \sqrt{1 + 7568}}{-2}
   \]
   \[
   x = \frac{-1 \pm \sqrt{7569}}{-2}
   \]
   \[
   x = \frac{-1 \pm 87}{-2}
   \]
   This gives two solutions:
   \[
   x = \frac{-1 + 87}{-2} = -43 \quad \text{(not valid since \( x \) must be non-negative)}
   \]
   \[
   x = \frac{-1 - 87}{-2} = 44
   \]

7. **Verify the solution:**
   Substitute \( x = 44 \) back into the original equation to ensure it satisfies the condition:
   \[
   \frac{(56-44)(55-44)}{2} + (56 - 44) \cdot 44 = \frac{12 \cdot 11}{2} + 12 \cdot 44 = 66 + 528 = 594
   \]
   The equation holds true.

Conclusion:
The maximum number of lines that could have the same slope is \( \boxed{44} \).