The conjecture is correct. Super-brief sketch proof: divide the grid into 3x3 boxes, aggregate those into horizontal and vertical strips of width 1 box = 3 squares; say a strip is "cheap" if it contains only 2k shaded squares (hence, exactly two per box). Working box by box from one end of a cheap strip to the other in both directions we see that the shaded squares must be in the middle transverse row/column of each box in such a strip. Hence there can't be both a cheap horizontal strip and a cheap vertical strip, because the box where they meet is overconstrained. So either all h-strips are not-cheap or all v-strips are not-cheap, and either of those gives you at least 2k2+k shaded squares.
(This, along with everything M.S. wrote, generalizes straightforwardly to arbitrary rectangular grids.)
(This, along with everything M.S. wrote, generalizes straightforwardly to arbitrary rectangular grids.)
on /blog/114