, 2024

*Proof.* The maximum achievable value of $N$ is $17$.

Let us denote the numbers on the two cards selected by Player $A$ in the $k$-th round as $a_k$ and $b_k$, where $a_k < b_k$. The number selected by Player $B$ is denoted as $c_k$, and the discarded number as $d_k$.

Let $T = d_1 + d_2 + d_3 + d_4$. Then we have $S + T = 1 + 2 + \dots + 8 = 36$.

Player $B$ has a strategy to ensure $S - T \ge -3$, and consequently $S \ge 17$ (note that $S$ must be an integer).

* If in the first round, $b_1 - a_1 \ge 4$, then Player $B$ selects $b_1$, and in the second round selects $a_2$. This gives $c_1 - d_1 \ge 4$ and $c_2 - d_2 \ge 1 - 8 = -7$.
* If in the first round, $b_1 - a_1 \le 3$, then Player $B$ selects $a_1$, and in the second round selects $b_2$. This gives $c_1 - d_1 \ge -3$ and $c_2 - d_2 \ge 1$.

Therefore, Player $B$ can always ensure $(c_1 + c_2) - (d_1 + d_2) \ge -3$.

In the third and fourth rounds, Player $B$ can always ensure $(c_3 + c_4) - (d_3 + d_4) \ge 0$. This is because after Player $A$ has chosen $a_3$ and $b_3$, $a_4$ and $b_4$ are also determined. Player $B$ can then select the pair $(a_3, b_4)$ or $(b_3, a_4)$ that yields the larger sum. Thus, Player $B$ can always guarantee $S - T \ge -3$.

Player $A$ has a strategy to ensure $S \le 17$:

* In the first round, Player $A$ selects $(3, 6)$. If Player $B$ chooses $3$, then in the second round Player $A$ selects $(4, 5)$. The sum of Player $B$'s selections in the first two rounds will not exceed $8$.

In the last two rounds, Player $A$ selects $(1, 2)$ and $(7, 8)$, and Player $B$'s selections in these rounds will not exceed $9$. Therefore, $S \le 17$.

* If in the first round Player $B$ chooses $6$, then in the second round Player $A$ selects $(1, 8)$, forcing Player $B$ to choose $1$. In the last two rounds, Player $A$ selects $(2, 4)$ and $(5, 7)$, and Player $B$'s selections in these rounds will not exceed $9$. Thus, $S \le 6 + 1 + 9 = 16$.

In conclusion, Player $A$ has a strategy to ensure $S \le 17$. $\Box$