Find all polynomials $P$ such that
$P(x) + \binom{2018}{2}P(x+2)+...+\binom{2018}{2106}P(x+2016)+P(x+2018)=$
$=\binom{2018}{1}P(x+1)+\binom{2018}{3}P(x+3)+...+\binom{2018}{2105}P(x+2015)+\binom{2018}{2107}P(x+2017)$ 
for all real numbers $x$.
1. We start by considering the given problem:
   \[
   P(x) + \binom{2018}{2}P(x+2) + \cdots + \binom{2018}{2106}P(x+2016) + P(x+2018) = \binom{2018}{1}P(x+1) + \binom{2018}{3}P(x+3) + \cdots + \binom{2018}{2105}P(x+2015) + \binom{2018}{2107}P(x+2017)
   \]
   for all real numbers \( x \).

2. To solve this, we consider a more general problem:
   \[
   \sum_{i=0}^n \binom{n}{i} (-1)^i P(x+i) = 0
   \]
   for all real numbers \( x \).

3. **Proof by Induction:**
   - **Base Case:** When \( n = 1 \), if \( P \) is a constant function, then:
     \[
     \sum_{i=0}^1 \binom{1}{i} (-1)^i P(x+i) = P(x) - P(x+1) = 0
     \]
     This holds true for any constant polynomial \( P \).

   - **Inductive Step:** Assume the statement holds for \( n = k-1 \). We need to show it holds for \( n = k \).

     If \( P \) is a polynomial of degree \( k-1 \), then \( P(x) - P(x+1) \) is a polynomial of degree \( k-2 \). We have:
     \[
     \sum_{i=0}^k \binom{k}{i} (-1)^i P(x+i) = \sum_{i=0}^k \left( \binom{k-1}{i} + \binom{k-1}{i-1} \right) (-1)^i P(x+i)
     \]
     This can be split into two sums:
     \[
     = \sum_{i=0}^{k-1} \binom{k-1}{i} (-1)^i P(x+i) + \sum_{i=1}^k \binom{k-1}{i-1} (-1)^i P(x+i)
     \]
     Shifting the index in the second sum:
     \[
     = \sum_{i=0}^{k-1} \binom{k-1}{i} (-1)^i P(x+i) + \sum_{i=0}^{k-1} \binom{k-1}{i} (-1)^{i+1} P(x+i+1)
     \]
     Combining the sums:
     \[
     = \sum_{i=0}^{k-1} \binom{k-1}{i} (-1)^i (P(x+i) - P(x+i+1))
     \]
     Since \( P(x) - P(x+1) \) is a polynomial of degree \( k-2 \), by the inductive hypothesis, this sum is zero. Therefore, the induction holds.

4. **Uniqueness:**
   - Suppose \( P \) has a degree greater than \( n-1 \). Let \( Q(x) \) be the unique polynomial of degree at most \( n-1 \) such that \( Q(k) = P(k) \) for all \( k \in \{1, 2, \ldots, n\} \).
   - Let \( R(x) = P(x) - Q(x) \). Then \( R \) is a nonzero polynomial of degree greater than \( n-1 \) and \( R(k) = 0 \) for all \( k \in \{1, 2, \ldots, n\} \).

5. We have:
   \[
   \sum_{i=0}^n \binom{n}{i} (-1)^i P(x+i) = 0
   \]
   Substituting \( P(x) = R(x) + Q(x) \):
   \[
   \sum_{i=0}^n \binom{n}{i} (-1)^i (R(x+i) + Q(x+i)) = 0
   \]
   Since \( Q(x) \) is a polynomial of degree at most \( n-1 \):
   \[
   \sum_{i=0}^n \binom{n}{i} (-1)^i R(x+i) = 0
   \]
   This implies:
   \[
   R(x+n) = -\sum_{i=0}^{n-1} \binom{n}{i} (-1)^i R(x+i)
   \]
   Using induction on \( x \), we can show that \( R(x) = 0 \) for all positive integers \( x \), implying \( R \) is the zero polynomial, which is a contradiction. Therefore, \( P \) can only have degree at most \( n-1 \).

The final answer is all polynomials \( \boxed{ P } \) with degree at most \( 2017 \).