Determine all pairs of natural numbers $(m, r)$ with $2014 \ge m \ge r \ge 1$  that fulfill
 $\binom{2014}{m}+\binom{m}{r}=\binom{2014}{r}+\binom{2014-r}{m-r} $
1. We start with the given equation:
   \[
   \binom{2014}{m} + \binom{m}{r} = \binom{2014}{r} + \binom{2014-r}{m-r}
   \]

2. We know that the binomial coefficient $\binom{n}{k}$ represents the number of ways to choose $k$ elements from a set of $n$ elements. Therefore, the equation can be interpreted combinatorially.

3. Consider the combinatorial identity:
   \[
   \binom{2014}{m} \binom{m}{r} = \binom{2014}{r} \binom{2014-r}{m-r}
   \]
   This identity holds because both sides count the number of ways to choose a group of $r$ elements and then a group of $m-r$ elements from a set of $2014$ elements.

4. Given that both sides of the equation are equal, we can infer that:
   \[
   \binom{2014}{m} + \binom{m}{r} = \binom{2014}{r} + \binom{2014-r}{m-r}
   \]
   This implies that the sum of the binomial coefficients on both sides must be equal.

5. We need to find pairs $(m, r)$ such that the above equation holds. We consider the cases where the binomial coefficients are equal:
   - Case 1: $\binom{2014}{m} = \binom{2014}{r}$
   - Case 2: $\binom{m}{r} = \binom{2014}{r}$

6. For Case 1:
   \[
   \binom{2014}{m} = \binom{2014}{r}
   \]
   This implies that $m = r$ or $m = 2014 - r$ because $\binom{n}{k} = \binom{n}{n-k}$.

7. For Case 2:
   \[
   \binom{m}{r} = \binom{2014}{r}
   \]
   This implies that $m = 2014$ because $\binom{2014}{r}$ is the binomial coefficient for choosing $r$ elements from $2014$ elements, and $\binom{m}{r}$ can only equal this if $m = 2014$.

8. Combining the results from both cases, we get the following pairs:
   - $(m, r) = (i, i)$ for $1 \le i \le 2014$
   - $(m, r) = (2014 - j, j)$ for $1 \le j \le 1006$
   - $(m, r) = (2014, k)$ for $1 \le k \le 2013$

The final answer is $\boxed{(i, i) \text{ for } 1 \le i \le 2014, (2014 - j, j) \text{ for } 1 \le j \le 1006, \text{ and } (2014, k) \text{ for } 1 \le k \le 2013}$.