10.6. On the board, 2011 numbers are written. It turns out that the sum of any three written numbers is also a written number. What is the smallest number of zeros that can be among these numbers?
Answer: 2009.

Solution. Let $n=2011$. Arrange the written numbers in non-decreasing order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}$. Since the number $a_{1}+a_{2}+a_{3}$ is written, then $a_{1}+a_{2}+a_{3} \geqslant a_{1}$, hence $a_{2}+a_{3} \geqslant 0$. Similarly, we get $a_{n-2}+a_{n-1}+a_{n} \leqslant a_{n}$, hence $a_{n-2}+a_{n-1} \leqslant 0$. Therefore, $0 \geqslant a_{n-2}+a_{n-1} \geqslant a_{2}+a_{3} \geqslant 0$; thus, $a_{2}+a_{3}=a_{n-2}+a_{n-3}=0$. Since $a_{2} \leqslant a_{3} \leqslant a_{n-2} \leqslant a_{n-1}$, it follows that $a_{2}=a_{3}=a_{n-2}=a_{n-1}$, and therefore, $a_{2}=a_{3}=\ldots=a_{n-1}=0$. Thus, among the written numbers, there are at least 2009 zeros.

An example with 2009 zeros and the numbers $1, -1$ shows that there can be exactly 2009 zeros.

Comment. Only the answer - 0 points.

Provided an example showing that there can be exactly $2009-1$ points.

Proved only that there must be at least 2009 zeros - 5 points.

Only shown that $a_{2}+a_{3} \geqslant 0$ (or $\left.a_{n-1}+a_{n-2} \leqslant 0\right)-$ 1 point.

Only shown that among the written numbers, there are no more than two positive and/or no more than two negative - 3 points.