11.6. The sequence of numbers $a_{1}, a_{2}, \ldots, a_{2022}$ is such that $a_{n}-a_{k} \geqslant$ $\geqslant n^{3}-k^{3}$ for any $n$ and $k$ such that $1 \leqslant n \leqslant 2022$ and $1 \leqslant k \leqslant$ $\leqslant 2022$. Moreover, $a_{1011}=0$. What values can $a_{2022}$ take?

(N. Agakhanov)
Answer. $a_{2022}=2022^{3}-1011^{3}=7 \cdot 1011^{3}$.

Solution. Writing the condition for $n=2022, k=1011$ and for $n=1011, k=2022$, we get

$$
a_{2022}=a_{2022}-a_{1011} \geqslant 2022^{3}-1011^{3}
$$

and

$$
-a_{2022}=a_{1011}-a_{2022} \geqslant 1011^{3}-2022^{3}
$$

that is, $a_{2022} \geqslant 2022^{3}-1011^{3} \geqslant a_{2022}$. This is where the answer follows from.

Remark. A sequence satisfying the condition exists, namely $a_{n}=n^{3}-1011^{3}$. Moreover, it is not difficult to show, similarly to the above solution, that such a sequence is unique. However, for solving the problem, it is not necessary to find all such sequences or even to provide an example of such a sequence.

Comment. It is proven that $a_{2022}=2022^{3}-1011^{3}-$ 7 points.

It is proven only that $a_{2022} \leqslant 2022^{3}-1011^{3}$ (or only $\left.a_{2022} \geqslant 2022^{3}-1011^{3}\right)-1$ point.