Let $f: N \to N$ be a function satisfying
(a) $1\le f(x)-x \le 2019$ $\forall x \in N$
(b) $f(f(x))\equiv x$ (mod $2019$) $\forall x \in N$
Show that $\exists x \in N$ such that $f^k(x)=x+2019 k, \forall k \in N$
Yep. Note that $2019|f(f(x))-f(x)+f(x)-x \le 2*2019$. Hence, for a single $x_0$ we either have $f(f(x_0))-x_0=2019$ or $f(f(x_0))-x_0=2*2019$. Note that if the second holds true for at least one such $x_0$ we can inductively get the required relation. Hence, assume that $f(f(x))=x+2019$ for each $x$. Note that $f$ is injective and thus reversible. Set $x=f^{-1}(1)$ to obtain $f(1)=f^{-1}(1)+2019\ge 2020$. However, $x+2019= f(f(x)) \ge f(x)+1$ $\Leftrightarrow$ $f(x) \le 2018+x$ for each $x$. Setting $x=1$ we get $f(1) \le 2019$ contradiction. (Note that it is well known that $f(f(x))=x+n$ where $n$ is odd yields no solutions for $f$.