Find all differentiable functions $f:\mathbb{R}\to\mathbb{R}$ such that
\[f'(x)=\frac{f(x+n)-f(x)}n\]
for all real numbers $x$ and all positive integers $n.$
1. We start with the given functional equation:
   \[
   f'(x) = \frac{f(x+n) - f(x)}{n}
   \]
   for all real numbers \( x \) and all positive integers \( n \).

2. First, we consider the case when \( n = 1 \):
   \[
   f'(x) = f(x+1) - f(x)
   \]
   This implies that the derivative of \( f \) at \( x \) is equal to the difference between \( f(x+1) \) and \( f(x) \).

3. Next, we differentiate both sides of the equation \( f'(x) = f(x+1) - f(x) \) with respect to \( x \):
   \[
   f''(x) = f'(x+1) - f'(x)
   \]
   Since \( f'(x) = f'(x+1) \) (as shown in the initial solution), we have:
   \[
   f''(x) = 0
   \]
   This implies that \( f'(x) \) is a constant function.

4. Let \( f'(x) = c \) for some constant \( c \). Integrating \( f'(x) = c \) with respect to \( x \), we get:
   \[
   f(x) = cx + d
   \]
   where \( d \) is an integration constant.

5. To verify that \( f(x) = cx + d \) satisfies the original functional equation, we substitute \( f(x) = cx + d \) into the equation:
   \[
   f'(x) = \frac{f(x+n) - f(x)}{n}
   \]
   Calculating the right-hand side:
   \[
   \frac{f(x+n) - f(x)}{n} = \frac{c(x+n) + d - (cx + d)}{n} = \frac{cx + cn + d - cx - d}{n} = \frac{cn}{n} = c
   \]
   Since \( f'(x) = c \), the equation holds true.

6. Therefore, the general solution to the functional equation is:
   \[
   f(x) = cx + d
   \]
   where \( c \) and \( d \) are constants.

The final answer is \( \boxed{ f(x) = ax + b } \) where \( a, b \in \mathbb{R} \).