Find all injective functions $f: \mathbb R \rightarrow \mathbb R$ such that for every real number $x$ and every positive integer $n$,$$ \left|\sum_{i=1}^n i\left(f(x+i+1)-f(f(x+i))\right)\right|<2016$$

[i](Macedonia)[/i]
1. We start with the given inequality for the function \( f: \mathbb{R} \rightarrow \mathbb{R} \):
   \[
   \left|\sum_{i=1}^n i\left(f(x+i+1)-f(f(x+i))\right)\right| < 2016
   \]
   This implies:
   \[
   -2016 < \sum_{i=1}^n i\left(f(x+i+1) - f(f(x+i))\right) < 2016
   \]

2. Consider the sum for \( n+1 \):
   \[
   \sum_{i=1}^{n+1} i\left(f(x+i+1) - f(f(x+i))\right)
   \]
   This can be written as:
   \[
   \sum_{i=1}^n i\left(f(x+i+1) - f(f(x+i))\right) + (n+1)\left(f(x+n+2) - f(f(x+n+1))\right)
   \]
   Using the given inequality, we have:
   \[
   -2016 < \sum_{i=1}^n i\left(f(x+i+1) - f(f(x+i))\right) < 2016
   \]
   Adding \((n+1)\left(f(x+n+2) - f(f(x+n+1))\right)\) to each part of the inequality, we get:
   \[
   -2016 + (n+1)\left(f(x+n+2) - f(f(x+n+1))\right) < \sum_{i=1}^{n+1} i\left(f(x+i+1) - f(f(x+i))\right) < 2016 + (n+1)\left(f(x+n+2) - f(f(x+n+1))\right)
   \]
   Since the sum \(\sum_{i=1}^{n+1} i\left(f(x+i+1) - f(f(x+i))\right)\) must also satisfy the original inequality, we have:
   \[
   -2016 < -2016 + (n+1)\left(f(x+n+2) - f(f(x+n+1))\right) < 2016
   \]
   and
   \[
   -2016 < 2016 + (n+1)\left(f(x+n+2) - f(f(x+n+1))\right) < 2016
   \]

3. Simplifying these inequalities, we get:
   \[
   -4032 < (n+1)\left(f(x+n+2) - f(f(x+n+1))\right) < 4032
   \]
   for all \( x \in \mathbb{R} \) and \( n \in \mathbb{Z}^+ \).

4. Suppose there exists \( t \) such that \( f(t+1) - f(f(t)) \neq 0 \). For sufficiently large \( m \in \mathbb{Z}^+ \), we have:
   \[
   |(m+1)(f((t-m-1)+m+2) - f(f((t-m-1)+m+1)))| = |(m+1)(f(t+1) - f(f(t)))| > 4032
   \]
   This contradicts the inequality \( -4032 < (m+1)(f(t+1) - f(f(t))) < 4032 \). Therefore, we must have:
   \[
   f(t+1) = f(f(t))
   \]
   for all \( t \in \mathbb{R} \).

5. Since \( f \) is injective, \( f(t+1) = f(f(t)) \) implies \( t+1 = f(t) \). Therefore, the function \( f \) must be:
   \[
   f(x) = x + 1
   \]
   for all \( x \in \mathbb{R} \).

The final answer is \( \boxed{ f(x) = x + 1 } \).