Replace $2021$ with $n\ge 4$ . We will prove that Alice draws at least $n-3$ diagonals by induction over $n$. Definition: Call a vertex strange if the number written on it is greater (or smaller) than both of the numbers written on its neighbors. Claim 1: If none of the vertices is strange, then there are more than $n-3$ diagonals. Proof: Assume that none of the vertices is strange. Let the vertices be $1,2,\cdots ,n$ in this order. Let the number written on the vertex $i$ be $a_i$. WLOG $a_1\ge a_2$. Since $a_2$ is not strange, we have $a_2\ge a_3$. Then, $a_3\ge a_4$. Similarly, $a_i\ge a_{i+1}$ for all $i=1,2,\cdots ,n$ where $a_{n+1}=a_1$. Then, $a_1\ge a_2\ge \cdots \ge a_n\ge a_1$. Hence, $a_1=a_2=\cdots =a_n$. Thus, Alice will draw $\frac{n(n-3)}2>n-3$ diagonals. Claim 2: Let $A$ be a strange vertex with neighbors $B$ and $C$. Then the numbers written on $B$ and $C$ differ by at most $1$. Proof: Let the numbers written on $A, B$ and $C$ be $a, b$ and $c$, respectively. Since $A$ is strange, we have either $a\leq b, c\leq a+1$ or $a\ge b, c\ge a-1$. Hence, $|b-c|\leq 1$, as desired. Now let's move on to induction. For $n=4$, if none of the vertices is strange, then by Claim 1, there are more than $n-3\ge 1$ diagonals. If there is at least $1$ strange vertex, then by Claim 2, Alice draws at least $1$ diagonals, as desired. Assume that for $n=k$ where $k\ge 4$, Alice draws at least $k-3$ diagonals. Let's prove that for $n=k+1$, Alice draws at least $k-2$ diagonals. If none of the vertices is strange, then by Claim 1, there are more than $k-2$ diagonals. Assume that $A$ is a strange vertex. Let the vertices be $A, B_1, B_2,\cdots ,B_k$ where $B_1$ and $B_k$ are neighbors of $A$. By Claim 2, the numbers written on $B_1$ and $B_k$ differ by at most $1$. Now remove the vertex $A$. By the induction hypothesis for the vertices $B_1, B_2,\cdots, B_k$, we find that there are at least $k-3$ diagonals. But, $B_1B_k$ is not one of these $k-3$ diagonals. Hence, in the initial diagram, there are at least $k-3+1=k-2$ diagonals, as desired. Finally, let's give an example for $2018$ diagonals. Label the vertices as follows.

I'll prove that the answer is 2018. Consider the following generalisation of the problem with the same rules but for a convex $n$-gon and define $f(n)$ to be the smallest number of diagonals that Alice can draw, for $n\geq 3$. First I'll prove that she has to draw at least 2018 diagonals. Take an $n$-gon that satisfies the conditions and let $A$ be a vertex of it containing the minimal real number(there could be more than one such vertex but this isn't a problem). Let $B$ and $C$ be the consecutive vertices to $A$ and let the real numbers in $A$,$B$ and $C$ be $a,b$ and $c$, respectively. We have that $a+1\geq b\geq a$ and $a+1\geq c\geq a$, thus $|b-c|\leq 1$ and Alice should draw the diagonal connecting vertices $B$ and $C$. Now consider the convex $(n-1)$-gon that consists of our starting $n$-gon without vertex $A$. It clearly satisfies the conditions, thus Alice should draw at least $f(n-1)$ diagonals inside of it. Hence we get that Alice has to draw at least $f(n-1)+1$ diagonals in the $n$-gon (the $f(n-1)$ diagonals inside of the $(n-1)$-gon and the diagonal connecting $B$ and $C$). Since $f(3)=0$, because there are no diagonals in a triangle, we get that $$f(2021)\geq f(2020)+1\geq \cdots \geq f(3) + 2018 = 2018$$and the desired bound is proven. Now I'll show that there does exists a regular $2021$-gon with the desired conditions in which Alice has to draw exactly $2018$ diagonals. Consider the following example: The smallest numbers in vertices are $1$ and $1011$, on one of the sides of the $2021$-gon between $1$ and $1011$ there are the numbers $2,3,4 \cdots 1010$; and on the other side there are the numbers $1.5;2.5; \cdots 1010.5$ consecutively. Clearly there are $2021$ real numbers and also if we check carefully, each of the $1009$ vertices with numbers from $2,3$ to $1010$ will be connected by Alice with two other vertices, namely , if $n={2,3,\cdots ,1010}$ it would be connected with $n-0.5$ and $n+0.5$. It's easy to see that there are no other diagonals marked by Alice, making up for a total of $2.1009=2018$.