We have $-2016<\sum_{i=1}^{n}{i(f(x+i+1)-f(f(x+i)))} <2016$ and also $$-2016<\sum_{i=1}^{n}{i(f(x+i+1)-f(f(x+i)))} +(n+1)(f(x+n+2)-f(f(x+n+1)))<2016.$$This gives $-4032<(n+1)(f(x+n+2)-f(f(x+n+1))) <4032$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}^+$. If there exists $t$ such that $f(t+1)-f(f(t))\neq 0$, for sufficiently large $m\in \mathbb{Z}^+$ we get $$|(m+1)(f((t-m-1)+m+2)-f(f((t-m-1)+m+1)))| =|(m+1)(f(t+1)-f(f(t)))|>4032.\quad \perp$$ Therefore $f(t+1)=f(f(t))$ for all $t\in \mathbb{R}$. And so $f(x)=x+1$ for all $x\in \mathbb{R}$ by injectivity.

The problem was proposed by FYROM

I have attached a document which contains the solution.

Same question was also asked in IMC I don't quite remember the year though

*Proposed by North Macedonia*

Let $$ g ( x ) = f ( x + 1) - f ( f ( x ) ) $$by the problems statemnet $ \forall x \in \mathbb{R} \text { and every positice integer n:} $ $$ \left |\sum_{ i = 1 }^n i \cdot g(x + i) \right| = \left| \sum_{ i = 1 }^n i \left ( f (x + i + 1) - f ( ( x + i ) ) \right ) \right | < 2016 $$then $ \forall x \in \mathbb{R} \text { and every positice integer } n > 1$ \begin{align*} \left |\sum_{ i = 1 }^{n} i \cdot g(x + i) \right | &< 2016 \\ | n \cdot g (x + n) | &\leq \left |-\sum_{ i = 1 }^{n - 1} i \cdot g(x + i) \right | + \left |\sum_{ i = 1 }^{n - 1} i \cdot g(x + i) + n \cdot g (x + n)\right | = \\ &= \left |\sum_{ i = 1 }^{n - 1} i \cdot g(x + i) \right | + \left |\sum_{ i = 1 }^{n} i \cdot g(x + i) \right | < 2\cdot 2016 \end{align*}$\text{let } x = y - n \text{ then } \forall y \in \mathbb{R} \text { and every positice integer } n > 1: $ $$ | g (y) | < \frac {2 \cdot 2016} {n} $$as n goes to infinity: $$ \lim_{n \to \infty} \frac {2 \cdot 2016} {n} = 0 $$so: \begin{align*} g(y)= 0 &= f ( y + 1) - f ( f ( y ) ) \text{ } \forall y \in \mathbb{R}\\ f ( y + 1 ) &= f ( f (y ) ) \text{ by remembering that } f \text{ is surjective} \\ f ( y ) &= y + 1 \text{ } \forall y \in \mathbb{R} \end{align*}checking the answer yields: $$ \left | \sum_{i = 1}^n i \left ( f (x + i + 1) - f ( f ( x + i ) ) \right ) \right | = \left | \sum_{i = 1}^n i \left (x + i + 2 - (x + i + 2) \right ) \right | = \left | \sum_{i = 1}^n i \cdot 0 \right | = 0 < 2016 $$so we are done!

Raunii wrote: Same question was also asked in IMC I don't quite remember the year though I was just about to write that, it's 1999.

cute!