Determine the smallest integer $n$ such that there exist $n$ real numbers $x_{1}, \ldots, x_{n}$ all belonging to the interval $(-1,1)$ and for which

$$
x_{1}+\cdots+x_{n}=0 \quad \text { and } \quad x_{1}^{2}+\cdots+x_{n}^{2}=2020
$$
The exercise requires finding the smallest integer $n$ satisfying a certain property. To show that the largest integer sought is an integer $c$, there are therefore necessarily two distinct parts: the analysis, in which it is established that any integer $n$ satisfying the stated property satisfies $n \geqslant c$, and the construction, in which an example of $c$ real numbers $x_{1}, \ldots, x_{c}$ from the interval $]-1,1[$ is given, satisfying the two mentioned equalities.

Analysis: Let $n$ be the smallest integer satisfying the property stated in the problem and let $x_{1}, \ldots, x_{n}$ be real numbers from the interval $]-1,1[$ for which

$$
x_{1}+\ldots+x_{n}=0 \quad \text { and } \quad x_{1}^{2}+\ldots+x_{n}^{2}=2020
$$

First, since $x_{i}^{2}-1$ for all real $i \leqslant k$, we know that

$$
0=x_{1}+x_{2}+\ldots+x_{n}>(-k)+x_{k+1}+x_{k+2}+\ldots+x_{n}
$$

Furthermore, for all $i \leqslant k$, the real number $x_{i}$ belongs to the interval $]-1,0[$, so $x_{i}^{2} \leq 1$, which implies $k \geqslant 1011$. Thus, for any non-zero real numbers satisfying the two equalities, there are at least 1011 strictly negative real numbers. Moreover, if $(x_{1}, \ldots, x_{n})$ is an $n$-tuple satisfying the two equalities in the problem, the $n$-tuple $(-x_{1}, \ldots, -x_{n})$ also satisfies the two equalities in the problem. Therefore, there are also at least 1011 strictly positive real numbers. In total, there are at least 2022 real numbers, so $n \geqslant 2022$.

Construction: Conversely, we show that $n=2022$ satisfies the property stated in the problem. The simplest way to do this is to look for a 2022-tuple of the form $(x, -x, x, -x, \ldots, x, -x)$ with $x$ well chosen. By substituting this 2022-tuple into the right equation, we get $2022 x^{2} = 2020$, so

$$
x = \pm \sqrt{\frac{2020}{2022}}
$$

By construction, for such an $x$, the 2022-tuple $(x, -x, x, -x, \ldots, x, -x)$ satisfies both equalities.

In conclusion, the smallest integer $n$ satisfying the property stated in the problem is $n=2022$.

Comment from the graders: The problem was very difficult and was solved by only a handful of students. While it was difficult to solve the problem completely, some parts of the proof were nevertheless accessible and could earn many partial points, such as providing a construction for 2022 or showing that $n > 2022$. Thus, even if very few students managed to provide a correct reasoning that proves there are no 2021 real numbers satisfying the two equations, most students who tackled the problem saw these two parts and earned valuable partial points. Being able to scrape partial points without necessarily finding the solution to a problem is an important asset for the olympiads. The key to the problem was to separate the positive real numbers from the negative real numbers and then establish inequalities between the different quantities (using simple upper bounds like $x^{2} \leq |x|$ or even $x_{1} + \ldots + x_{k} < k$) to finally combine them correctly. There were therefore several opportunities to earn additional partial points by establishing one or the other of the intermediate inequalities. We detail here the reasoning encountered and why they do not allow for a conclusion.

- We recall that the real numbers belonged to the interval $]-1; 1[$, meaning they could not take the value $\pm 1$. It is regrettable to see that several students did not take this important detail into account.
- Several students claim that a good way to proceed is to pair the real numbers with their opposites, and that therefore $n$ is odd. But of course, just because a practical way to find real numbers is to choose numbers and their opposites does not mean there is no better way to do it and in particular that there is no solution in the case $n=2021$. In the same vein, some students claim that in the case where $n=2021$, there must be three real numbers with a sum of zero, but this is obviously not true.
- Many students give very vague reasoning in which the real numbers are taken to the limit at $\pm 1$. But here, we are given an integer $n$ and we assume that there is a solution to the system of two equations with $n$ unknowns, so it does not really make sense to take the real numbers to the limit.