Is it true that for every $n\ge 2021$ there exist $n$ integer numbers such that the square of each number is equal to the sum of all other numbers, and not all the numbers are equal? (O. Rudenko)
1. Let $\{x_i\}_{i=1}^n$ be the set of $n$ integers such that the square of each number is equal to the sum of all other numbers. Define $S = \sum_{i=1}^n x_i$.
2. The given condition translates to $x_i^2 = S - x_i$ for each $i$. Rearranging, we get $x_i^2 + x_i - S = 0$.
3. This is a quadratic equation in $x_i$, and its solutions are given by the quadratic formula:
   \[
   x_i = \frac{-1 \pm \sqrt{1 + 4S}}{2}
   \]
4. Since $x_i$ must be an integer, $1 + 4S$ must be a perfect square. Let $1 + 4S = k^2$ for some integer $k$. Then $S = \frac{k^2 - 1}{4}$.
5. The solutions to the quadratic equation are:
   \[
   x_i = \frac{-1 + k}{2} \quad \text{and} \quad x_i = \frac{-1 - k}{2}
   \]
6. Let $a = \frac{-1 + k}{2}$ and $b = \frac{-1 - k}{2}$. These are the two possible values for $x_i$. Note that $a$ and $b$ are integers.
7. Suppose we have $m$ numbers equal to $a$ and $n-m$ numbers equal to $b$. Then the sum $S$ can be written as:
   \[
   S = ma + (n-m)b
   \]
8. Substituting $a$ and $b$ into the equation for $S$, we get:
   \[
   S = m\left(\frac{-1 + k}{2}\right) + (n-m)\left(\frac{-1 - k}{2}\right)
   \]
9. Simplifying, we find:
   \[
   S = \frac{m(-1 + k) + (n-m)(-1 - k)}{2} = \frac{-m + mk - n + m - nk + mk}{2} = \frac{(m - n)k - n}{2}
   \]
10. Equating this to the earlier expression for $S$, we have:
    \[
    \frac{(m - n)k - n}{2} = \frac{k^2 - 1}{4}
    \]
11. Multiplying through by 4 to clear the denominators, we get:
    \[
    2(m - n)k - 2n = k^2 - 1
    \]
12. Rearranging, we obtain:
    \[
    k^2 - 2(m - n)k + 2n - 1 = 0
    \]
13. This is a quadratic equation in $k$. For $k$ to be an integer, the discriminant must be a perfect square. The discriminant is:
    \[
    \Delta = [2(m - n)]^2 - 4(2n - 1) = 4(m - n)^2 - 8n + 4
    \]
14. Simplifying, we get:
    \[
    \Delta = 4(m - n)^2 - 8n + 4
    \]
15. For $\Delta$ to be a perfect square, $4(m - n)^2 - 8n + 4$ must be a perfect square. This is a necessary condition for the existence of such integers $x_i$.
16. If $2n - 1$ is a prime number, then $2k + 1$ must be one of $\{-2n + 1, -1, 1, 2n - 1\}$.
17. Checking each case:
    - $2k + 1 = -2n + 1$ implies $k = -n$, which does not fit.
    - $2k + 1 = -1$ implies $k = -1$, which does not fit.
    - $2k + 1 = 1$ implies $k = 0$, which does not fit.
    - $2k + 1 = 2n - 1$ implies $k = n - 1$, which does not fit.

Therefore, there are no such numbers when $2n - 1$ is prime.

The final answer is $\boxed{\text{False}}$