https://aops.com/community/p11366699 programjames1 wrote: The second condition gives $f(x)\le x$. Now suppose for the sake of contradiction that $f(x)=f(y)$ when $x\ne y$. Then, $$x+f(x)=x+f(y)|f(y+f(x))\le y+f(x)\implies x\le y$$But also $$y+f(y)=y+f(x)|f(x+f(y))\le x+f(y)\implies y\le x$$And so $x=y$, a contradiction. Therefore, as $f$ is from $\mathbb{N}\to\mathbb{N}$, and $x+1\ge f(x+1)\ne f(x)$, an easy induction proves that $f(x)=x$ is the only solution.

For $x\ge 2018$ we have $f(x)\le x$ from the second condition. Assume that $x, y\ge 2017$. Then we have $x+f(y) \le f(y+f(x)) \le y+f(x)$. Swapping $x$ and $y$ implies $x+f(y)=y+f(x)$. $\textbf{Claim:}$ We have $f(p+2017)=p+2017$ for any prime $p>2017$. Proof: From the second condition we have $f(p+2017) \in \{2016,2018,p+2017\}$. If $f(p+2017) \neq p+2017$. Putting $x=p+2017, y=p+2017-f(p+2017)$ gives contradiction. Putting $y=p+2017$ in $x+f(y)=y+f(x)$ implies $f(x)=x$ for $x\ge 2017$. Assume $x$ is large enough, $y<2017$ in first condition. We have $x+f(y)|x+y \implies f(y)\le y$ for all $y<2017$. Assume $x<2017$ in the second condition. We have $|f(x)-2017|\le |x-2017| \implies x\le f(x) \implies x=f(x)$ for $x<2017$. Hence $f(x)=x$ for all $x$.$\square$ $\textbf{REMARK:}$ $p>2017$ is necessary in the claim because we can have $f(p+2017)=2017-p$ for $p<2017$.

Very nice problem: Hint:$$p\ge f(2018) f(p+2017)=p+2017$$$$y\ge 2017,f(2018) $$$$f(y)=y$$All natural numbers $f(x)=x$