\section*{Problem 1 - 141231}

The numbers 1, 2, ..., 64 are entered into the 64 squares of a chessboard such that in the first horizontal row from left to right the numbers \(1,2, \ldots, 8\), in the second horizontal row from left to right the numbers \(9,10, \ldots, 16\) and so on are placed in this order.

Someone is now to place eight rooks on the squares of the chessboard such that no two of them can attack each other.

Then he should form the sum \(S\) of the numbers on the squares occupied by the rooks. All possible values of \(S\) are to be given.

Note:

Two rooks can attack each other if and only if they stand on a common horizontal or vertical row of squares.
}

If we start from the final arrangement of the 8 rooks, then in each row there must be exactly one of the 8 rooks, and the same applies to the columns. The numbers of the placed rooks are therefore in ascending order of the form

\[
8(k-1)+s_{k}, k=1,2, \ldots, 8
\]

where here the column numbers \(s_{1}, s_{2}, \ldots, s_{8}\) must satisfy the condition

\[
\left\{s_{1}, s_{2}, \ldots, s_{8}\right\}=\{1,2, \ldots, 8\}
\]

which is clearly necessary for 8 rooks and is also sufficient to ensure that no two rooks can attack each other. From this, due to the commutativity of addition, it immediately follows that

\[
s_{1}+s_{2}+\ldots+s_{8}=1+2+\ldots+8=\frac{8 \cdot 9}{2}=36
\]

and thus

\[
S=\sum_{k=1}^{8}\left(8(k-1)+s_{k}\right)=9 \sum_{k=1}^{8} k-8 \cdot 8=9 \cdot 36-64=260
\]

i.e., \(S\) has, independently of the arrangement of the rooks, as long as it only meets the conditions of the problem here, always the same value 260.