The answer is
$$\begin{aligned} N &:= \left( \sum_{k=1}^{n} 2k \right)^2 + \left( \sum_{k=1}^{n} (2k-1) \right)^2 - \sum_{k=1}^{n} (2k-1)(2k) \\ &= n^2(n+1)^2 + n^4 - \left( \frac{2n(n+1)(2n+1)}{3} - n(n+1) \right) \\ &= \frac{n(6n^3 + 2n^2 + 1)}{3} \end{aligned}$$

Denote
$X = a_1 + a_2 + \cdots + a_n,$
$Y = b_1 + b_2 + \cdots + b_n,$
$S = X^2 + Y^2,$
$T = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n.$

Claim. $T \le 1 \cdot 2 + 3 \cdot 4 + \cdots + (2n-1)(2n)$

Proof. Since there are finitely many possible ways to assign the values, $T$ must attain a maximum value. For this maximal case, we may assume $a_i < b_i$ for each $i$. Note that we may assume $\{a_i\}$ and $\{b_i\}$ are increasing by the rearrangement inequality.

Under these assumptions, we must have $a_1 = 1$. We are going to show that $a_i = 2i - 1$ and $b_i = 2i$ for all $i$. Suppose on the contrary that this does not hold. Then there is a smallest index $k$ such that $b_k \neq 2k$. Note that we must have $a_i = 2i - 1$ for $i = 1, 2, \dots, k$ and $a_{k+1} = 2k$ (clearly $k \neq n$). This implies $a_k < a_{k+1} < b_k < b_{k+1}$. By the rearrangement inequality, we have
$$a_k b_k + b_{k+1} a_{k+1} < a_k a_{k+1} + b_{k+1} b_k.$$
This means we can swap $b_k$ and $a_{k+1}$ to increase the value of $T$, contradiction.

Therefore, $T$ is maximized when $a_i = 2i - 1$ and $b_i = 2i$ for all $i$. This proves the claim. $\square$

Now, we show that $A$ can guarantee to obtain score $N$. A strategy is to choose the card with the largest label in each round. Note that $a_i \ge 2(n+1-i)$ for each $i$ since at least one of the cards $2(n+1-i), 2(n+1-i)+1, \dots, 2n$ is not chosen in the $i$th round. As $X+Y = 1+2+\dots+2n$ is fixed and $X > Y$ (because of this strategy), $S$ is smaller when $X$ is smaller. Thus,
$$S \ge (2+4+\dots+2n)^2 + (1+3+\dots+(2n-1))^2.$$
Together with the claim, we deduce
$$S - T \ge \left( \sum_{k=1}^{n} 2k \right)^2 + \left( \sum_{k=1}^{n} (2k-1) \right)^2 - \sum_{k=1}^{n} (2k-1)(2k) = N$$
as desired.

Next, we show that $B$ can guarantee that the score of $A$ is at most $N$. A strategy is to pair up the cards with labels $2m-1$ and $2m$ for each $m = 1, 2, \dots, n$, and choose a card in the same pair as the card just chosen by $A$ in each round. It is easy to see that
$$T = 1 \cdot 2 + 3 \cdot 4 + \dots + (2n-1)(2n).$$
Also, as $|a_i - b_i| = 1$ for each $i$, we have
$$\begin{aligned} S &= X^2 + Y^2 = \frac{(X+Y)^2 + (X-Y)^2}{2} \\ &= \frac{1}{2} \left[ (1+2+\dots+2n)^2 + \left( \sum_{i=1}^{n} (a_i - b_i) \right)^2 \right] \\ &\le \frac{(1+2+\dots+2n)^2 + n^2}{2} \end{aligned}$$

Equality holds when $a_i - b_i = 1$ for all $i$ or $a_i - b_i = -1$ for all $i$. This is exactly the case when
$$S-T = \left(\sum_{k=1}^{n} 2k\right)^2 + \left(\sum_{k=1}^{n} (2k-1)\right)^2 - \sum_{k=1}^{n} (2k-1)(2k) = N.$$
The proof is complete.