For which real numbers $c$ is there a straight line that intersects the curve
\[ y = x^4 + 9x^3 + cx^2 + 9x + 4\] 
in four distinct points?
To determine for which real numbers \( c \) there exists a straight line that intersects the curve 
\[ y = x^4 + 9x^3 + cx^2 + 9x + 4 \]
in four distinct points, we need to analyze the behavior of the polynomial function \( f(x) = x^4 + 9x^3 + cx^2 + 9x + 4 \).

1. **Find the derivative of \( f(x) \):**
   \[
   f'(x) = \frac{d}{dx}(x^4 + 9x^3 + cx^2 + 9x + 4) = 4x^3 + 27x^2 + 2cx + 9
   \]

2. **Determine the conditions for \( f'(x) \) to have three real roots:**
   For a straight line to intersect the curve at four distinct points, \( f(x) \) must have three critical points (two local minima and one local maximum or vice versa). This means \( f'(x) \) must have three real roots.

3. **Analyze the second derivative \( f''(x) \):**
   \[
   f''(x) = \frac{d}{dx}(4x^3 + 27x^2 + 2cx + 9) = 12x^2 + 54x + 2c
   \]
   For \( f'(x) \) to have three real roots, \( f''(x) \) must have two real roots (one minimum and one maximum).

4. **Solve for the roots of \( f''(x) = 0 \):**
   \[
   12x^2 + 54x + 2c = 0
   \]
   Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{-54 \pm \sqrt{54^2 - 4 \cdot 12 \cdot 2c}}{2 \cdot 12} = \frac{-54 \pm \sqrt{2916 - 96c}}{24}
   \]
   For the roots to be real, the discriminant must be non-negative:
   \[
   2916 - 96c \geq 0
   \]
   Solving for \( c \):
   \[
   2916 \geq 96c \implies c \leq \frac{2916}{96} = \frac{243}{8}
   \]

5. **Conclusion:**
   The value of \( c \) must be less than or equal to \(\frac{243}{8}\) for the polynomial \( f(x) \) to have the required critical points, ensuring that a straight line can intersect the curve at four distinct points.

The final answer is \( \boxed{ c \leq \frac{243}{8} } \)