combinatorics
We are given an $18*18$ table, all of whose cells may be black or white. Initially all the cells are colored white. We may perform the following operation:
Choose one column or one row and change the colour of all cells in this column or row. Is it possible by repeating the operation to obtain a table with exactly $16$ black cells?
I know that this question is based on invariant principle but I am not getting the invariant.
Let $V$ be the vector space over the field with two elements, $\Bbb F_2$, generated by all matrices of shape $N\times N$, $N=18$, which have exactly one row, or exactly one column with one entries, all other entries being zero. We consider the following map from $V$ to $\Bbb F_2$:
$X$ in $V$ is mapped to $$ f(X)= (x_{11}+x_{22}+\dots+x_{N-1,N-1}+x_{N,N}) + (x_{12}+x_{23}+\dots+x_{N-1,N}+x_{N,1})\ . $$ (Sum on two consecutive fixed diagonals of $X$, where the indices are considered modulo $N$.)
Then each generator of $V$ is mapped to zero, since the "bits" taken in $f(X)$ are exactly two in each row and/or column. So $f$ vanishes on $V$. But it does not vanish on any quadratic submatrix $A=A(I)$ with ones exactly on the positions $I\times I$, $I$ being a proper interval of $\{1,2,\dots,N\}$. (For instance $I=\{1,\dots,16\}$ for our $N=18$.)
So in our case the answer to the problem is negative. (The "change of the color on a line/columns" corresponds to adding a generator of $V$ to a matrix. We are starting with a zero matrix. The matrices that can be obtained are exactly those in the vector space $V$ of dimension $N+N-1$.)
Note: I can provide some simple computer check in sage, if needed.
Later edit: It is simpler to test the equations $$ x_{is}-x_{js}-x_{it}+x_{jt}=0\ ,\qquad 1\le i<j\le N\ ,\ 1\le s<t\le N\ , $$ that can be extracted from all $2\times 2$ minors of a matrix $X$ in order to test/decide if a matrix $X$ can be realized. (The minus is also a plus, but makes it simpler to see that e.g. each of the above "one row matrix of ones" satisfy the equation, explicitly $1-1-0+0=0$, and correspondingly for the column matrices in the base of $V$...)
Here is a solution for $n=12$, that you can adapt for any (odd or even) $n>4$. Draw a checkerboard, but make sure the four inner corners stay white.
Best Answer
First notice that the order in which we perform the row/column operations doesn't matter and that using two times the same column/row does nothing. Now suppose we can make $16$ black cells using $r$ row-moves and $c$ column-moves. Imagine to color the columns and rows we used, every coloured cell is now black, except the cells coloured 'two times' that are white. The total number of coloured cells is $$18(r+c)-rc.$$ The number of cells coloured two times are $rc$. So the total number of black cells are $$18(r+c)-2rc.$$ Now we have to solve (for $0\le r,c\le18$) $$18(r+c)-2rc=16$$ that can be rewritten as $$r=9-\frac{73}{9-c}.$$ Since $r$ must be a integer, we have that $(9-c) \mid 73$, the only possibility is $c=8$, but we would have $r<0$. Hence there are no solutions.
EDIT: As pointed out by @filipos we must consider also the case $c=10$, which gives $r>18$ and so it has to be discarded.