Does there exist a function $ f $ such that $ f ( x ) $ is an integer for only finitely many values of $ x $? Consider $ f : \mathbb R \to \mathbb R $ such that $ f ( x ) \le f ( y ) $ whenever $ x \le y $ and $ f ^ { 2018 } ( z ) \in \mathbb Z \ \forall z \in \mathbb R $. Does there exist a function $ f $ such that $ f ( x ) $ is an integer for only finitely many values of $ x $?
I think that the condition of the problem forces $ f ^ n ( x ) \in \mathbb Z $ for all integer $ n \ge 2018 $. But this doesn't help I guess.
And I'm somewhat sure that the answer is NO.
Side-Note: $ f ^ n ( x ) $ denotes $ n $-th composition of $ f $.
 A: The answer is indeed "no", as you expected. The number $ 2018 $ and the set $ \mathbb Z $ are almost irrelevant. Also, the assumptions are somehow more than necessary. So I present a more general argument, which in my own opinion, explains the situation more clearly.
Let $ f : \mathbb R \to \mathbb R $, and for each $ n \in \mathbb N $, denote range of $ f ^ n $ by $ A _ n $. It's easy to inductively see that for each $ m , n \in \mathbb N $ with $ m \ge n $, we have $ A _ m \subseteq A _ n $.
Assume that $ f $ is increasing, and there is an $ N \in \mathbb N $ such that $ A _ N $ is finite, say with $ M $ elements. Then for each $ n \in \mathbb N $ with $ n \ge N $, $ A _ n $ is a subset of $ A _ N $, and thus finite. Hence, for every $ n \in \mathbb N $ we can consider $ a _ n $, the maximum element of $ A _ { N + n } $. Since $ f $ is increasing, it can inductively be shown that $ f ( a _ n ) = a _ { n + 1 } $ and $ a _ { n + 1 } \le a _ n $, for every $ n \in \mathbb N $. Now, since $ A _ N $ has $ M $ elements and $ a _ n \in A _ { N + n } \subseteq A _ N $ for every $ n \in \mathbb N $, by pigeonhole principle, there are $ m , n \in \{ 0 , \dots , M \} $ with $ m < n $ and $ a _ m = a _ n $. This implies that letting $ b = a _ { N + M - 1 } $, we have $ f ( b ) = b $.
Now, since $ f $ is increasing, for every $ x \in \mathbb R $ with $ x \ge b $, it can inductively be shown that $ f ^ n ( x ) \ge b $, for each $ n \in \mathbb N $. Also, since $ b = a _ { N + M - 1 } $, we have $ f ^ { N + M - 1 } ( x ) \le b $, which together with the previous result, shows that $ f ^ { N + M - 1 } ( x ) = b $.
At last, note that since $ f $ is increasing, if there is an $ x \in \mathbb R $ with $ x > b $ and $ f ( x ) = b $, then for every $ y \in \mathbb R $ with $ b \le y \le x $ we have $ f ( y ) = b $, which shows that there are infinitely many (in fact, uncountably many) $ y \in \mathbb R $ with $ f ( y ) = b $. Thus, if we require that this happens only for finitely many $ y $ (or even for countably many $ y $), then for every $ x \in \mathbb R $ with $ x > b $, we must have $ f ( x ) \ne b $. We combine this with the previous paragraph to inductively prove that for every $ x \in \mathbb R $ with $ x > b $ and every $ n \in \mathbb N $ with $ n < N + M $, one must have $ f ^ { N + M - n - 1 } ( x ) = b $. By the previous paragraph, the statement holds for $ n = 0 $. Assuming the statement for $ n = m $ where $ m + 1 < N + M $, note that if we had $ f ^ { N + M - ( m + 1 ) - 1 } ( x ) > b $, we would have $ f ^ { N + M - m - 1 } ( x ) \ne b $, which contradicts the assumption. Thus we must have $ f ^ { N + M - ( m + 1 ) - 1 } ( x ) \le b $. Also, by the previous paragraph, we have $ f ^ { N + M - ( m + 1 ) - 1 } ( x ) \ge b $, which proves the statement for $ n = m + 1 $. At last, as the result holds for $ n = N + M - 1 $, we must have $ x = b $, which is absurd.
To sum up, we've proven that if $ f : \mathbb R \to \mathbb R $ is increasing and there is an $ N \in \mathbb N $ such that $ A _ N $ is finite, then there is a $ b \in A _ N $ such that $ \{ x \in \mathbb R | f ( x ) = b \} $ is not finite, or even countable.
Going back to the original problem, if we let $ N = 2018 $, then since $ A _ N \subseteq \mathbb Z $, $ A _ N \subseteq A _ 1 $ and $ A _ 1 \cap \mathbb Z $ is finite, therefore $ A _ N $ is finite. Hence there must be a $ b \in A _ N \subseteq \mathbb Z $ with infinitely many $ x \in \mathbb R $ satisfying $ f ( x ) = b $, contradictiong the assumption of the problem. Thus there is no function $ f $ satisfying all those conditions.
