Find all injective functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x \in \mathbb{R}$ and all $n \in \mathbb{N}:$

$$
\left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\right| \leqslant 2019
$$
Let $x$ be a real number and let $m \geqslant 1$ be an integer. We consider the following difference:

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

This difference is always equal to $|m(f((x-m)+m+1)-f(f((x-m)+m)))|$. By applying the hypothesis of the statement for $n=m$ and $n=m-1$, and using the fact that for all real numbers $x, y$ we have $|x-y| \leqslant|x|+|y|$, we deduce that:

$$
|m(f(x+1)-f(f(x)))| \leqslant 2019+2019
$$

In particular, for any integer $m \geqslant 1: |(f(x+i+1)-f(f(x+i)))| \leqslant \frac{4038}{m}$. We conclude that for any real number $x: f(x+1)=f(f(x))$. Since $f$ is assumed to be injective, for any real number $x$ we have that $f(x)=x+1$.