Hang on... why are there two people in the question... is the question wrong or something? If not isnt it just $2k-1$ by colouring alternate squares and the bound is trivial?

gghx wrote: Hang on... why are there two people in the question... is the question wrong or something? If not isnt it just $2k-1$ by colouring alternate squares and the bound is trivial? Yes I think so haha or we're just both confused. Anyways does anyone have any idea about what this competition is, the "2nd Final Mathematical Cup Junior/Senior Division 2020", like where did it take place and was it an international competition (because the problems are on the easier side)?

credit to ok1985. $\color{blue} \textbf{\text{Answer.}}$ $\boxed{n=4k}$. $\color{blue} \textbf{\text{Bound.}}$ We claim that we must have that $n\geqslant 4k$. For an even $n$, just divide the grid into $2\times 2$ squares and note that in every such square has at most $1$ black square, therefore we must have $n\geqslant 4k$. Thus, let us fix $k$ and consider still that same $n$. Now, biggest $\text{odd}\times\text{odd}$ grid, which is smaller than $n$ is $2m-1$, where $n=2m$. So let us consider $(2m-1)\times (2m-1)$ grid and we must have in each row and column, $k$ black squares. Let us divide this $(2m-1)\times (2m-1)$ grid into two parts: $(2m-2)\times (2m-2)$ grid and we are left with one column and row. Since $2m-2$ is even, we must have at most $\frac{(2m-2)^2}{4}=(m-1)^2$ black squares in $(m-1)\times (m-1)$ grid. In the remaining row and column, there are in total at most $2k=2\cdot\frac{n}{4}=m$ black squares. In total in $(2m-1)\times (2m-1)$ grid, we have $(2m-1)\cdot k=\frac{(2m-1)\cdot n}{4}=\frac{(2m-1)\cdot m}{2}$ black squares. But note that $$ (m-1)^2+m<\frac{(2m-1)\cdot m}{2} \Longleftrightarrow m^2-m+1<m^2-\frac{m}{2}\Longleftrightarrow 2<m,$$which implies a contradiction. If $m=1,2$ then obviously two squares touch each other. Since $n>1$ we conclude that we must have for fixed $k$, $n$ bigger than or equal to $4k$. $\color{blue} \textbf{\text{Construction.}}$ Consider $4\times 4$ square and colour grid as shown. For $4k\times 4k$ put $4\times 4$ square exactly to middle and cover each quadrant with $2\times 2$ squares, what are similar to the square in the middle $4\times 4$ square in that quadrant. [asy][asy]size(5cm); path R = box( (-1,2), (3,6) ); path S = box( (-3,0), (5,8) );fill(S, white); fill(shift(-1,3)*unitsquare, black); fill(shift(0,5)*unitsquare, black); fill(shift(2,4)*unitsquare, black); fill(shift(1,2)*unitsquare, black); fill(shift(1,0)*unitsquare, grey); fill(shift(3,0)*unitsquare, grey); fill(shift(3,2)*unitsquare, grey); fill(shift(4,6)*unitsquare, grey); fill(shift(4,4)*unitsquare, grey); fill(shift(2,6)*unitsquare, grey); fill(shift(0,7)*unitsquare, grey); fill(shift(-2,7)*unitsquare, grey); fill(shift(-2,5)*unitsquare, grey); fill(shift(-3,3)*unitsquare, grey); fill(shift(-3,1)*unitsquare, grey); fill(shift(-1,1)*unitsquare, grey); path K = box( (1,2), (1,6) );fill(K, white);draw(K, black+1.4); path L = box( (-1,4), (3,4) );fill(L, white);draw(L, black+1.4); for (int i=-3; i<=4; ++i) { draw(shift(i,7)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,6)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,5)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,4)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,3)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,2)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,1)*unitsquare); } for (int i=-3; i<=4; ++i) { draw(shift(i,0)*unitsquare); } draw(R, black+1.4);draw(S, black+1.4); [/asy][/asy]