Find the set of all real numbers $k$ with the following property: For any positive, differentiable function $f$ that satisfies $f^{\prime}(x) > f(x)$ for all $x,$ there is some number $N$ such that $f(x) > e^{kx}$ for all $x > N.$
1. **Given Condition**: We are given that for any positive, differentiable function \( f \) that satisfies \( f'(x) > f(x) \) for all \( x \), there exists some number \( N \) such that \( f(x) > e^{kx} \) for all \( x > N \).

2. **Choosing \( k < 1 \)**: Let's choose \( k < 1 \). We need to show that for such \( k \), the condition holds. Consider the function \( g(x) = f(x)e^{-x} \). We compute its derivative:
   \[
   g'(x) = \frac{d}{dx} \left( f(x)e^{-x} \right) = f'(x)e^{-x} - f(x)e^{-x} = (f'(x) - f(x))e^{-x}
   \]
   Given \( f'(x) > f(x) \), it follows that:
   \[
   g'(x) = (f'(x) - f(x))e^{-x} > 0
   \]
   Hence, \( g(x) \) is an increasing function. This implies:
   \[
   f(x)e^{-x} > f(0)e^{0} = f(0)
   \]
   Therefore:
   \[
   f(x) > f(0)e^{x}
   \]
   For some \( N \), it holds that \( f(0) > e^{(k-1)x} \) for \( x \ge N \). Thus:
   \[
   f(x) > e^{kx} \text{ for all } x \ge N
   \]

3. **Counterexample for \( k \ge 1 \)**: Now, we need to find a counterexample for \( k \ge 1 \). Consider the function:
   \[
   f(x) = 
   \begin{cases} 
   \frac{e^{3x}}{2} & \text{for } x \le 0 \\
   \frac{2e^x - e^{-x}}{2} & \text{for } x > 0 
   \end{cases}
   \]
   For \( x \le 0 \):
   \[
   f'(x) = \frac{3e^{3x}}{2} \quad \text{and} \quad f(x) = \frac{e^{3x}}{2}
   \]
   Thus:
   \[
   f'(x) = 3f(x) > f(x)
   \]
   For \( x > 0 \):
   \[
   f'(x) = \frac{2e^x + e^{-x}}{2} \quad \text{and} \quad f(x) = \frac{2e^x - e^{-x}}{2}
   \]
   Thus:
   \[
   f'(x) = \frac{2e^x + e^{-x}}{2} > \frac{2e^x - e^{-x}}{2} = f(x)
   \]
   Therefore, \( f'(x) > f(x) \) for all \( x \). However, for \( k \ge 1 \), we need to check if \( f(x) > e^{kx} \) for all \( x > N \). For \( x \le 0 \):
   \[
   f(x) = \frac{e^{3x}}{2} \quad \text{and} \quad e^{kx} \le e^x
   \]
   For \( x > 0 \):
   \[
   f(x) = \frac{2e^x - e^{-x}}{2} \quad \text{and} \quad e^{kx} \ge e^x
   \]
   Clearly, \( f(x) \) does not satisfy \( f(x) > e^{kx} \) for all \( x > N \) when \( k \ge 1 \).

The final answer is \( \boxed{ k < 1 } \)