61. (9th grade) In a square table consisting of $8 \times 8$ cells, natural numbers from 1 to 64 are arranged in order. A number is selected, and the row and column in which it is located are crossed out. From the remaining numbers, another number is selected, and the row and column in which it is located are crossed out again. This continues until all rows and columns are crossed out. Find the sum of the eight selected numbers.
61. The chosen number $N=M_{i}+8\left(n_{k}-1\right)$, where $M_{i}$ is the column number, and $n_{k}$ is the row number in which it stands. Then the total sum of the eight chosen numbers can be written as:

$$
S=\sum_{i=1, k=1}^{i=8, k=8} M_{i}+8\left(n_{k}-1\right)=\frac{8+1}{3} 8+8 \frac{7+1}{2} 7=260
$$