Since there are 17 rooks in 8 rows, at least one row contains at least 3 rooks. Let this be row A. Disregard row A in the following.

The remaining 7 rows house at least 17 - 8 = 9 rooks. The Pigeonhole principle implies the existence of a row with at least 2 rooks. Denote one of these rows B and remove it from further consideration.

There at at least 9 - 8 = 1 rooks in the remaining 6 rows. It follows there is at least one row with at least one rook. Denote one of these C.

Now, let R_C be any rook in rwo C. Since row B contains more than 1 rook, there is a rook R_B in a column different from that of R_C so that the two do not threaten each other.

Since row A contains at least 3 rooks, there is a rook R_A in columns different from those of R_B and R_C. So chosen rooks do not threaten each other.

(The problem is a simplification of Problem 1.4.5 from Mathematics as Problem Solving by A. Soifer.)