Outer Floor - Inner Floor.

Part 1. Proof that the degree of $f(x)$ cannot be more than 1.

We are going to prove this by contradiction. Assume that the degree of some solution $f(x)$ is greater than or equal to 2, and also that $n$ is an arbitrary integer number. Let us consider any real number $x$ satisfying the condition that $n\leq x < n+1.$ Then the functional equation given for $f(x)$ implies, $f(n) \leq f(x) < f(n)+1.$ Using the continuity of the polynomial and taking limits in this inequality as $x \rightarrow (n+1)^-,$ then $f(n)\leq f(n+1) \leq f(n)+1.$ This implies that $0\leq f(n+1) -f(n) \leq 1\quad\quad\quad\quad (*)$ Now, we can see that this leads to a contradiction, using the Mean Value Theorem, $f(n+1)-f(n) = f ' (\theta_n),$ where $\theta_n \in (n, n+1).$ Since $f ' (x)$ is a polynomial whose degree is at least 1, then $\lim_{n \rightarrow \infty} (f(n+1)- f(n)) = \lim_{n \rightarrow \infty} f ' (\theta_n) =\pm \infty,$ which contradicts $(*).$ Therefore, Part 1 is proven.

Part 2. Proof of the fact that such a polynomial $f(x)$ can be only of form $f(x)=b$ or $f(x)=x+b,$ where $b$ is an integer satisfying that $1\leq b \leq 2018.$

Since the degree of $f(x)$ can be 0 or 1, then we can assume that $f(x)=ax+b$ for certain values of $a$ and $b$ . It is easy to see that $b$ has to be integer, because $b=f(0)$ and from the fact, that follows from the functional equation given in the problem, that $f(0)= \left\lfloor f(0)\right \rfloor.$ Additionally, the condition $1\leq f(0)\leq 2018$ implies that $1\leq b \leq 2018$

Now let us prove that $a$ can be only 0 or 1. The functional equation given in the problem implies that $a\left\lfloor x\right \rfloor + b \leq ax+b <a \left\lfloor x\right \rfloor +b+1$ This inequality implies $0\leq a(x-\left\lfloor x\right \rfloor ) <1$ Then, assuming that $x$ is not integer, $0\leq a <\frac{1}{x-\left\lfloor x\right \rfloor }$ Making $x$ tend to any integer number from the left, we obtain that $0\leq a \leq 1.$ From the fact that $a \left\lfloor x\right \rfloor +b$ is always integer for any values of $x$ due to the equation $a \left\lfloor x\right \rfloor +b =\left\lfloor ax+b\right \rfloor,$ then $a$ has to be integer too. Therefore, $a$ can be only 0 or 1. So the Part 2 is proven.

Part 3. Finding the number of possible functions $f(x)$ satisfying the conditions of the problem.

According to the Part 2 , there can be 2018 constant functions satisfying those conditions and 2018 more functions of the form $f(x)=x+b$ that satisfy the conditions also. So the number of possible functions will be $\boxed{4036}.$

Let's start with some obvious observations: If $a$ is an integer, then so is $f(a)$ since $\lfloor f(a) \rfloor=f(\lfloor a \rfloor) =f(a)$ . Furthermore, $f$ maps the interval $[a,a+1)$ into the interval $[f(a),f(a)+1)$ : If $x\in[a,a+1)$ then $\lfloor f(x) \rfloor=f(\lfloor x \rfloor) =f(a)$ . Thus, by continuity, $f(a+1)$ must be either $f(a)$ or $f(a)+1$ . The Mean Value Theorem now tells us that there are infinitely many $x$ -values where $f'(x)$ is 1 or 0. With $f'(x)$ being a polynomial, this implies that $f'(x)$ is either constant 0 or constant 1. This leaves only the functions $f(x)=k$ and $f(x)=x+k$ where $k$ is an integer between 1 and 2018, a total of $\boxed{4036}$ functions.

Again, a very charming problem! Thank you!