9.2. On the board, there are $n$ different integers, any two of which differ by at least 10. The sum of the squares of the three largest of them is less than three million. The sum of the squares of the three smallest of them is also less than three million. For what largest $n$ is this possible?
Answer. For $n=202$.

Solution. Notice immediately that $990^{2}+1000^{2}+1010^{2}=(1000-10)^{2}+1000^{2}+(1000+10)^{2}=3 \cdot 1000^{2}+2 \cdot 10^{2}$, which is greater than three million. On the other hand, $989^{2}+999^{2}+1009^{2}=(1000-11)^{2}+(1000-1)^{2}+(1000+9)^{2}=3 \cdot 1000^{2}-6 \cdot 1000+(9^{2}+1^{2}+11^{2})$, which is less than three million.

If there are not fewer than 102 non-negative numbers on the board, then the three largest of them are not less than $99 \cdot 10=990$, $100 \cdot 10=1000$, and $101 \cdot 10=1010$ respectively. Then, by the above observation, the sum of their squares is greater than three million, which is impossible. Therefore, there are no more than 101 non-negative numbers on the board. Similarly, there are no more than 101 negative numbers. Thus, the total number of numbers is no more than $101+101=202$.

An example for $n=202$ is given by the numbers $-1009, -999, -989, \ldots, -9$, $9, 19, \ldots, 999, 1009$. From the above observation, this example works.

Remark. Other examples that work include, for instance, the numbers $-1005, -995, -985, \ldots, -5, 5, 15, \ldots, 995, 1005$.

Comment. Only the proof that $n \leqslant 202$ is worth 5 points.

Only a correct example for $n=202$ (without justification of optimality or with incorrect justification) is worth 2 points. A correct example with $n=201$ numbers does not add points.

It may happen that a participant provides a correct example and a correct proof of optimality for exactly this example, but incorrectly counts the numbers in the example (for example, by counting that the sequence $5, 15, 25, \ldots, 1005$ contains exactly 100 numbers). In this case, 1 point should be deducted for the arithmetic error. (If this error leads to the participant incorrectly proving the estimate $n \leqslant 200$, no points are added for this part).

Works may be encountered where the authors believe that all numbers on the board are positive (or non-negative). In this case, the following criteria apply.

Only the proof that there are no more than $101$ positive (non-negative) numbers is worth 3 points.

An example consisting of non-negative numbers in this case does not add points.