We will prove that for any $n$ the maximum number of lines $f(n)=n+3$ (for a rectangle not square). We will use strong induction to prove $f(n) \le n+3$. Obvious for $n=1,2$. Assume that it is true for $1,2,3,...,n-1$. If there is only one rectangle which has an edge on the left edge of main rectangle, apply induction to get $f(n) \le f(1)+f(n-1)-3 \le n+3$. If there are at least two choose one at the bottom and call it $A$. Draw the line which is determined by the top side of $A$. Assume that "this line" intersects $a$ rectangles and partitons main rectangle into $X, Y$ big rectangles (we now have $2a$ rectangles). Apply induction to big rectangles to get $f(n) \le f(X)+f(Y)-a-3 \le X+Y-a+3=n+3$ (there are $a$ left lines of these $a$ rectangles which is counted twice, "this line" and two sides of the main rectangle) For construction divide main rectangle into rows.