10.5. On the board, 10 natural numbers are written, some of which may be equal, and the square of each of them divides the sum of all the others. What is the maximum number of different numbers that can be among the written ones?
Answer. Four

Solution. An example for four different numbers: 1,1,1,2,2,3,5,5,5,5.

Let among the listed numbers there are exactly $n \geq 2$ different ones, the maximum of which we denote by $x$, and the sum of all numbers by $\mathrm{S}$. Then the sum of all numbers except the maximum does not exceed $(9-n) x + x - 1 + x - 2 + \ldots + x - (n-1) = 9x - \frac{n(n-1)}{2}$, which is not less than $x^2$, since it is divisible by it. Therefore, the inequality $x^2 - 9x + \frac{n(n-1)}{2} \leq 0$ has solutions in natural numbers, and $\frac{9 - \sqrt{81 - 2n(n-1)}}{2} \leq x \leq \frac{9 + \sqrt{81 - 2n(n-1)}}{2}$. The positivity of the discriminant gives us $n \leq 6$, and $x \leq 8, S - x \leq 72 - 1 = 71$. Consider each case of the maximum number separately.

a) $x=8$, then $S - x \leq 71$ and is divisible by 64, so $S = 72$. Then $72 - k$ is divisible by $k^2, k \leq 7$ only for $k=1$, so in this case $n \leq 2$.

b) $x=7$, then $S - x \leq 62$ and is divisible by 49, so $S = 56$. Then $56 - k$ is divisible by $k^2, k \leq 6$ only for $k=1$, so in this case $n \leq 2$.

c) $x=6$, then $S - x \leq 53$ and is divisible by 36, so $S = 42$. Then $42 - k$ is divisible by $k^2, k \leq 5$ only for $k=1,2$, so in this case $n \leq 3$.

d) $x=5$, then $S - x \leq 44$ and is divisible by 25, so $S = 30$. Then $30 - k$ is divisible by $k^2, k \leq 4$ for $k=1,2,3$, so in this case $n \leq 4$. The desired set must contain 10 natural numbers with a sum of 30, among which there must be $1,2,3,5$. Now it is not difficult to construct an example for $n=4: 1,1,1,2,2,3,5,5,5,5$. This example is not unique.

Grading criteria. It is proven that $n \leq 4: 5$ points. Example for $n=4: 2$ points. Estimate $n \leq 6: 2$ points.