Fix $k\ge 0$ integer. Considering $f(x) = x+k$ establishes $A\supseteq \{k:k\ge 1989\}$. We now show $A\subseteq \{k:k\ge 1989\}$. Fix an $f\in F$. We claim $f(n)\ge n$ for all $n\in\mathbb{N}$. Assume this is false, and $n_0$ be the smallest positive integer with $f(n_0)<n_0$. Set $f(n_0)=n_0-k$. We then have (a) $n_0\ge 2$ and (b) $1\le k\le n_0-1$. Now, inserting $n_0$ in place of $x$ we find that $f(n_0-k) = n_0-2k<n_0-k$. This contradicts with the minimality of $n_0$, thus $A\subseteq \{k:k\ge 1989\}$. Hence, $A=\{k:k\ge 1989\}$.