Problem 1

Find the largest integer $k$ having the following property: for any real numbers $x_{1}, x_{2}, \ldots, x_{2024}$ such that

$$
x_{1}^{2}=\left(x_{1}+x_{2}\right)^{2}=\cdots=\left(x_{1}+x_{2}+\cdots+x_{2024}\right)^{2}
$$

there are at least $k$ of them that are all equal.
Solution

First, note that if $x_{1}=1, x_{2 k}=-2$ for $k \in\{1, \ldots, 675\}, x_{2 k+1}=2$ for $k \in\{1, \ldots, 675\}$ and $x_{k}=0$ for $k \in\{1352, \ldots, 2024\}$, we verify that each of the sums is $\pm 1$, and there are no 676 real numbers all equal. Thus, $k \leqslant 675$.

Now, let's show that given 2024 real numbers satisfying the property, we can always find 675 that are all equal.

Let, for all $k \in\{1, \ldots, 2024\}, s_{k}=x_{1}+\cdots+x_{k}$. Then, for all such $k$, we have $\left|s_{k}\right|=\left|x_{1}\right|$. In particular, the numbers $s_{k}$ can take at most two possible values, namely $x_{1}$ and $-x_{1}$.

We deduce that for all $k \in\{2, \ldots, 2024\}, x_{k}=s_{k}-s_{k-1}$ can take at most three possible values, namely $-2 x_{1}, 0$ or $2 x_{1}$. But then, by the pigeonhole principle, there is one of these values that is taken by at least $\left\lceil\frac{2023}{3}\right\rceil=675$ real numbers.