Find all injective functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that for every real number \( x \) and every positive integer \( n \),

\[
\left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\right|<2016
\]
From the condition of the problem we get

$$
\left|\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$

Then

$$
\begin{aligned}
& |n(f(x+n+1)-f(f(x+n)))| \\
= & \left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))-\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right| \\
< & 2 \cdot 2016=4032
\end{aligned}
$$

implying

$$
|f(x+n+1)-f(f(x+n))|<\frac{4032}{n}
$$

for every real number $x$ and every positive integer $n$.
Let $y \in \mathbb{R}$ be arbitrary. Then there exists $x$ such that $y=x+n$. We obtain

$$
|f(y+1)-f(f(y))|<\frac{4032}{n}
$$

for every real number $y$ and every positive integer $n$. The last inequality holds for every positive integer $n$ from where $f(y+1)=f(f(y))$ for every $y \in \mathbb{R}$ and since the function $f$ is an injection, then $f(y)=y+1$. The function $f(y)=y+1$ satisfies the required condition.