Team Selection Test for IMO, 2024

Answer: $\lambda = 2^{2024}$.
Let $x_i = a_i - a_{i+1}$. Since for all real numbers
$$\max\{x, y\} = \frac{x+y}{2} + \left|\frac{x-y}{2}\right|$$
we have
$$b_i - b_{i+1} = x_i + \frac{x_{i+1}}{2} + \frac{x_{i+2}}{2} + \left| \frac{x_{i+1}}{2} - \frac{x_{i+2}}{2} \right| \quad (1)$$

Let $y_i = \frac{x_i}{2} + \left|\frac{x_i}{2}\right|$ and $z_i = \frac{x_i}{2} - \left|\frac{x_i}{2}\right|$. Then, $y_i = x_i$, $z_i = 0$ for $x_i \ge 0$ and $y_i = 0$, $z_i = x_i$ for $x_i < 0$. Inserting it to (1) we get
$$(b_i - b_{i+1})^{2024} = (x_i + y_{i+1} + z_{i+2})^{2024} \le 2^{2023} (x_i^{2024} + (y_{i+1} + z_{i+2})^{2024}) \quad (2)$$
by the Power Mean Inequality. Because $y_i$ is always a non-negative real number and $z_i$ is always a non-positive real number, we have
$$|y_{i+1} + z_{i+2}| \le |y_{i+1}| + |z_{i+2}|$$
which in turn implies that
$$(y_{i+1} + z_{i+2})^{2024} \le y_{i+1}^{2024} + z_{i+2}^{2024}$$

Since $y_i^{2024} + z_i^{2024} = x_i^{2024}$ for all indices $i$, by summing the inequalities (2) for all indices $i$ we get the desired inequality.
To prove that $\lambda = 2^{2024}$ is the best possible value, consider $n = 2m$ and choose the sequence as $\{a_i\} = \{1, 2, \dots, n-1, n, n-1, \dots, 1\}$. Then, $\lambda$ has to satisfy the condition
$$\lambda \cdot n \geq 2^{2024} (n-4) + 1 + 1 + 0 + 0$$
for every $n$. Taking $n$ to infinity, we see that $\lambda$ should be at least $2^{2024}$.