Korean Mathematical Olympiad Final Round

If $x_1 = x_2 = \dots = x_{2024} = y_1 = y_2 = \dots = y_{2024} = p$, then $p^2 \ge 1-p$, that is, $p \ge \frac{1+\sqrt{5}}{2}$. So, $p$ must be at least $\frac{1+\sqrt{5}}{2}$. We prove that the condition holds for $p = \frac{1+\sqrt{5}}{2}$, which implies that the answer is $\frac{1+\sqrt{5}}{2}$.
Let $p = \frac{1+\sqrt{5}}{2}$, and let
$$S = \{(x_1, \dots, x_{2024}) \mid 0 \le x_1 \le \dots \le x_{2024} \le 1, x_1 + x_2 + \dots + x_{2024} = 2024p\}.$$
For $X = (x_1, x_2, \dots, x_{2024})$, $Y = (y_1, y_2, \dots, y_{2024}) \in S$, we regard $x_0 = y_0 = 0$ for convenience, and let
$$f(X, Y) = \sum_{i=1}^{2024} x_i (y_{2025-i} - y_{2024-i}) \\ = x_1(y_{2024} - y_{2023}) + x_2(y_{2023} - y_{2022}) + \dots + x_{2023}(y_2 - y_1) + x_{2024}y_1.$$
Since $f(X, Y) = y_1(x_{2024} - x_{2023}) + y_2(x_{2023} - x_{2022}) + \dots + y_{2023}(x_2 - x_1) + y_{2024}x_1$, $f(X, Y)$ is symmetric with regard to $X$ and $Y$. If $x_1 = x_2 = \dots = x_{k_1} = a_1$, $x_{k_1+1} = \dots = x_{k_1+k_2} = a_2$, $\dots,$ $x_{k_1+k_2+\dots+k_{m-1}+1} = \dots = x_n = a_m$ for $0 \le a_1 < a_2 < \dots < a_m \le 1$, we denote $X$ by $(a_1^{k_1}, a_2^{k_2}, \dots, a_m^{k_m})$ where $k_m = 2024 - k_1 - k_2 - \dots - k_{m-1}$, and let $l(X) = m$. And in this case, we have
$$f(X, Y) = a_1(y_{2024} - y_{2024-k_1}) + a_2(y_{2024-k_1} - y_{2024-k_1-k_2}) + \dots + a_m y_{k_m} \\ = a_1 Y_1 + a_2 Y_2 + \dots + a_m Y_m$$
where $Y_t = y_{2024-k_1-k_2-\dots-k_{t-1}} - y_{2024-k_1-k_2-\dots-k_t} = y_{k_t+k_{t+1}+\dots+k_m} - y_{k_t+1+\dots+k_m}$ for $t = 1, 2, \dots, m$.
Now we consider $X, Y \in S$ such that $f(X, Y)$ is minimized, and subject to that $l(X) + l(Y)$ is minimum. Let $X = (a_1^{k_1}, a_2^{k_2}, \dots, a_m^{k_m})$.
Lemma. There is no $1 \le i < m$ such that $0 < a_i, a_{i+1} < 1$.

Proof. Suppose $0 < a_i < a_{i+1} < 1$ for some $i$.
Case 1. $\frac{Y_i}{k_i} \le \frac{Y_{i+1}}{k_{i+1}}$.
Let $a = \frac{k_i a_i + k_{i+1} a_{i+1}}{k_i + k_{i+1}}$. Then, $a_i < a < a_{i+1}$, and for
$$X' = (a_1^{k_1}, \dots, a_{i-1}^{k_{i-1}}, a_{i-1}^{k_{i-1}+k_{i+1}}, a_{i+2}^{k_{i+2}}, \dots, a_m^{k_m})$$
we have
$$f(X, Y) - f(X', Y) \ge 0$$
and $l(X') = m - 1 < l(X)$, which yields a contradiction to the fact that $l(X) + l(Y)$ is minimum.
Case 2. $\frac{Y_i}{k_i} > \frac{Y_{i+1}}{k_{i+1}}$.
Choose a positive real number $e$ such that $e < a_i - a_{i-1}$ (if $i = 1$, let $a_0 = 0$) and $a_{i+1} + \frac{k_i e}{k_{i+1}} < a_{i+2}$ (if $i = m - 1$, let $a_{m+1} = 1$). Let $a = a_i - e$ and $b = a_{i+1} + \frac{k_i e}{k_{i+1}}$, and consider $X' = (a_1^{k_1}, a_2^{k_2}, \dots, a_{i-1}^{k_{i-1}}, a^{k_i}, b^{k_{i+1}}, a_{i+2}^{k_{i+2}}, \dots, a_m^{k_m})$. Since $k_i a + k_{i+1} b = k_i a_i + k_{i+1} a_{i+1}$, $X' \in S$. However, $f(X, Y) - f(X, Y') > 0$, yielding a contradiction. Therefore there is no $1 \le i < m$ such that $0 < a_i, a_{i+1} < 1$. $\square$
The above lemma also holds for $Y$. Thus, there exist non-negative integers $k_1, k_2, k_3, s_1, s_2, s_3$ and real numbers $0 < a, b < 1$ such that
$$\bullet \quad k_1 + k_2 + k_3 = s_1 + s_2 + s_3 = 2024,$$
$$\bullet \quad x_1 = x_2 = \cdots = x_{k_1} = 0, x_{k_1+1} = \cdots = x_{k_1+k_2} = a, x_{k_1+k_2+1} = \cdots = x_{2024} = 1$$
$$\bullet \quad y_1 = y_2 = \cdots = y_{s_1} = 0, y_{s_1+1} = \cdots = y_{s_1+s_2} = a, y_{s_1+s_2+1} = \cdots = y_{2024} = 1$$
$$\bullet \quad k_2 a + k_3 = s_2 b + s_3 = 2024p.$$
We note that $k_1, k_2, k_3, s_1, s_2, s_3$ could be zero, and we have
$$f(X, Y) = a y_{k_2+k_3} + (1-a) y_{k_3} = b x_{s_2+s_3} + (1-b) x_{s_3}.$$
If $k_2 = 0$, then $x_1 + x_2 + \cdots + x_{2024} = k_3 \ne 2024p$ (since $k_3$ is an integer), so $k_2 > 0$. Thus, $k_3 < k_2 a + k_3 = 2024p$ and $k_1 = 2024 - (k_2 + k_3) < 2024 - (k_2 a + k_3) = 2024(1-p)$. That is,
$$k_1 < 2024(1-p), \quad k_2 > 0, \quad k_3 < 2024p,$$
and similarly,
$$s_1 < 2024(1-p), \quad s_2 > 0, \quad s_3 < 2024p.$$
Since $y_1 = y_2 = \dots = y_{s_1} = 0$, $y_{s_1+1} = \dots = y_{s_1+s_2} = a$, and $y_{s_1+s_2+1} = \dots = y_{2024} = 1$, in order for $f(X, Y) = a y_{k_2+k_3} + (1-a) y_{k_3}$ to be minimized, the following must hold.
(i) $y_{k_2+k_3+1} = \dots = y_{2024} = 1$
(ii) $y_{k_3+1} = \dots = y_{k_2+k_3}$
(iii) $y_1 = \dots = y_{k_3}$
Similarly, the following must hold.
(i') $x_{s_2+s_3+1} = \dots = x_{2024} = 1$
(ii') $x_{s_3+1} = \dots = x_{s_2+s_3}$
(iii') $x_1 = \dots = x_{s_3}$
By (i) and (i'), $k_1 \le s_3$ and $s_1 \le k_3$.
If $k_1 < s_3$, then $y_{k_3+1} = 1$ by (i) and (ii), so $k_1 + k_2 \le s_3 < 2024p$. Then by (iii), $k_3 = s_2$, $y_{k_3} = b$ and $s_1 = 0$, $k_1 + k_2 = s_3$. So,
$$\begin{aligned} a &= \frac{2024p - k_3}{k_2} = \frac{2024p - k_3}{2024 - k_1 - k_3} \ge \frac{2024p - k_3}{2024 - k_3} = 1 - \frac{2024(1 - p)}{2024 - k_3} \\ b &= \frac{2024p - s_3}{s_2} = \frac{2024p - k_1 - k_2}{k_3} = \frac{k_3 - 2024(1 - p)}{k_3} = 1 - \frac{2024(1 - p)}{k_3} \end{aligned}$$
Since $2024p > k_3 = 2024 - (k_1 + k_2) = 2024 - s_3 > 2024(1 - p)$, we have
$$\begin{aligned} f(X, Y) &= a y_{k_2+k_3} + (1-a) y_{k_3} = a + (1-a)b = 1 - (1-a)(1-b) \\ &= 1 - \frac{2024(1-p)}{2024-k_3} \cdot \frac{2024(1-p)}{k_3} \ge 1-p. \end{aligned}$$
So, we may assume that $k_1 = s_3$ and by the symmetry, we may further assume that $s_1 = k_3$. Then, $k_2 = s_2$, and
(i) $y_{k_2+k_3+1} = \dots = y_{2024} = 1$
(ii) $y_{k_3+1} = \dots = y_{k_2+k_3} = b$
(iii) $y_1 = \dots = y_{k_3} = 0$.