We call the squares removed from the chessboard "holes," and let the chessboard have $k$ holes in total.
As shown in Figure 7-6, the chessboard is divided into 9 regions $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, and $O$ by 4 straight lines.

By the conditions, $A \cup G$ has at least 2 holes, $E \cup O$ has at least 3 holes, $C \cup H$ has at least 2 holes, and $D \cup F$ has at least 3 holes, thus $k \geqslant 2+3+2+3=10$.

If $k=10$, then region $D$ has no holes. By symmetry, regions $A$, $B$, $C$, and $D$ have no holes. Therefore, since $A \cup G$ has at least 2 holes and $A$ has no holes, $G$ must have at least 2 holes. Similarly, $H$ must have at least 2 holes. Since $A \cup E$ has at least 3 holes and $A$ has no holes, $E$ must have at least 3 holes. Similarly, $F$ must have at least 3 holes. Thus, regions $E$, $F$, $G$, and $H$ together have at least 10 holes. Since the chessboard has exactly 10 holes, $E$ and $F$ each have 3 holes, and $G$ and $H$ each have 2 holes, and region $O$ has no holes.

Since $A \cup E$ has at least 1 hole in each column and $A$ has no holes, $E$ must have at least 1 hole in each column. Since $E$ has only 3 holes, each column in $E$ must have exactly 1 hole. Therefore, at least one of the squares 1 and 2 is not a hole, so one of the 2 lines in the figure must have 5 consecutive pieces, which is a contradiction. Therefore, $k \geqslant 11$.
When $k=11$, as shown in Figure 7-7 (a more natural construction than in the "scoring standard"), we use "knight's move" to place the pieces, ensuring that there are no 5 consecutive pieces on the same line in the chessboard.

In summary, $k_{\min }=11$.
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline 1 & & & & & 6 & & \\
\hline & & & 5 & & & & \\
\hline & 2 & & & & & 9 & \\
\hline & & & & 8 & & & \\
\hline & & 3 & & & & & 11 \\
\hline 7 & & & & & 10 & & \\
\hline & & & 4 & & & & \\
\hline
\end{tabular}

Figure 7-7
Example 9 In a $7 \times 8$ grid chessboard, each cell contains a chess piece. If two chess pieces are in cells that share a vertex, then these two pieces are said to be connected. Now, $r$ pieces are removed from these pieces, so that among the remaining pieces, there are no 5 pieces in a straight line (horizontal, vertical, or at $45^{\circ}$ or $135^{\circ}$) in sequence. Find the minimum value of $r$. (2007 National High School Mathematics Competition Problem)