This is a well known cops-robber game. For a graph $G$ the minimal number of cops needed to catch the robber is called the cops number of $G$, denoted as $c(G)$. In the case when $ G$ has diameter $2$, as in the this problem, there is a result of L. Lu, X. Peng, (On Meyniel’s conjecture of the cop number, Journal of Graph Theory, 2012) stating $$c(G)\leq 2\sqrt{n}-1$$An improvement by Z. Wagner (https://arxiv.org/pdf/1312.7555.pdf) shows $$c(G)\leq \sqrt{2n}$$Putting $n=2019$ gives $c(G)\leq 63$. So, the result here is even sharper. Fortunately, it could be proved, by exactly the same method as used in the link above, that $c(G)$ is less or equal than $\left\lceil k\right\rceil $ , where $k$ is the positive root of the equation $(k+2)(k+3)=2(n+4)$. For $n=2019$ we get $c(G)\leq 62$ as desired. I may post a full solution and an overview of the game in my blog soon, there is nothing involved.

Solution. Put $n$ instead of $2019$. Claim. $k=\left\lceil \sqrt{2n+8+1/4}-5/2\right\rceil$ cops are enough to capture the robber. For $n=2019$ it yields $k=62$. In order to prove the claim we need the following Lemma. Let $ G$ is a graph with diameter $ 2$ and $ H$ is a subgraph of $ G$ with maximal degree $ k$. We play a cop-robber like game, but the robber is restricted to move only along the edges of $ H$. Cops can move, as before, on the whole graph $ G$. Then, $ k$ cops are enough to catch the robber. Let after his turn the robber is at a vertex $ v_0$ of $ H$ and cops are at $ u_1,u_2,\dots,u_k$. Let $ v_1,v_2,\dots,v_{\ell}, \ell\leq k$ be the neighbours of $ v_0$ in $ H$. There is a path with at most two edges connecting $ u_i$ with $ v_i, i=1,2,\dots,\ell$. Hence with one move we can dispose all cops currently in $ u_i,i=1,2,\dots,\ell$ in some vertices adjacent correspondingly to $ v_1,v_2,\dots,v_\ell$. If the robber moves on the next turn , he will be caught. If he stays in $ v_0$, the cops move to $ v_1,v_2,\dots,v_{\ell}$ on the next turn and the robber is captured. $ \blacksquare$ Proof of the claim. If there is a vertex $ v$ with degree at most $ k$ we are done since we deploy the cops at the neighbours $ v_1,v_2,\dots,v_{\ell}, \ell\le k$ of $ v$. Indeed, if the robber is at vertex $ u$, either $ u$ is one among $ v,v_i,i=1,\dots,\ell$ or there exists a path $ v v_j u$. In either case the cops capture the robber in their next move. So, suppose all degrees of $ G$ are at least $ k+1$. We take a vertex $ v_0$ and place at it a stationary cop. Then remove $ v_0$ and all of its neighbours. The remaining graph $ H_1$ has at most $ n-k-2$ vertices and the robber is constrained to move only along its edges. If $ H_1$ has a vertex with degree at most $ k-1$ we apply Lemma 1 and and are done, otherwise all degrees of $ H_1$ are at least $ k$. Again we take $ v_1\in H_1$, place at it a stationary cop , delete $ v_1$ and all of its neighbours and obtain a subgraph $ H_2$. Continuing like that, at $ j$-th step we have a graph $ H_j$ with at most $ n-(k+2)-(k+1)-\dots -(k+3-j)$ vertices and $j$ cops are already exhausted. Thus, if it happens Lemma 1 was never applied, at $ k-1$-th step we are having a graph $ H_{k-1}$ with $ n-(k+2)-(k+1)-\dots-4$ vertices and only one cop left. It's enough $ H_{k-1}$ to be with at most $ 2$ vertices and the last cop would catch the robber. So, we are done if $ n-(k+2)-(k+1)-\dots-4\leq 2$. It's equivalent with $ (k+2)(k+3) - 2(n+4)\geq 0$. But $\sqrt{2n+8+1/4}-5/2$ is exactly the positive root of $ (x+2)(x+3) - 2(n+4)= 0 $. Hence, $k$ cops capture the robber. Here are additional comments and some history overview of the cop-robber game.

I guess $\sqrt{2n+8+1/4}-5/2$

Thanks! Edited.