Easy to induct on. For the base case, where the cards are $1$ and $2,$ Beri can clearly force the trinomial $x^2-2x+1.$ Assume it works for $n=k.$ We show it works for $n=k+2.$ Clearly, if Ari picks $k+1$ or $k+2,$ Beri can pick $k+2$ and $k+1,$ respectively and force the polynomial $x^2-(k+2)x+(k+1) = (x-(k+1))(x-1),$ which all have integer roots. This completes the induction, so we are done.

Claim: If Ari chooses $a$, thus Beri can choose $"a+1"$ or $"a-1"$. Proof: The equation $x^2-(a+1)x+a=0$ has solution for $x = \frac{a+1 - \sqrt{(a-1)^2}}{2} = 1$. And the equation $x^2-(a)x+a-1=0$ has root 1 also. Thus the claim is proved. Now, if Ari picks a odd number, just use the strategy for $a+1$ and if Ari picks a even number, use strategy $"a-1"$ and then is easy to see that Beri always wins.

Vloe wrote: Ari and Beri play a game using a deck of $2020$ cards with exactly one card with each number from $1$ to $2020$. Ari gets a card with a number $a$ and removes it from the deck. Beri sees the card, chooses another card from the deck with a number $b$ and removes it from the deck. Then Beri writes on the board exactly one of the trinomials $x^2-ax+b$ or $x^2-bx+a$ from his choice. This process continues until no cards are left on the deck. If at the end of the game every trinomial written on the board has integer solutions, Beri wins. Otherwise, Ari wins. Prove that Beri can always win, no matter how Ari plays. Consider the $1010$ sets $\mathcal{X}=\{(1,2),(3,4),\cdots (2019,2020)\}$,whatever Ari choses Beri choses the complimentary number in these sets,formally if Ari choses an odd number then Beri choses $b=a+1$ and if Ari choses an even number Beri choses $b=a-1$.Its easy to see this works.$\square$

Consider the following $1010$ ordered pairs: $$(1, 2); (3, 4); \ldots; (2019, 2020).$$Whenever Ari picks a card numbered $a$, Beri can just pick the card numbered $b$ such that $a$ and $b$ form one of the ordered pairs mentioned above. This clearly works, as the trinomial $$x^2 - (n+1)x + n = (x-1)(x-n)$$has two integer solutions. It follows that Beri can always win the game. $\blacksquare$