Nice question .Not so easy to explain but I will try. Lets look at the edges between two adjacent squares. There are $2n(n-1)$ of them. Call an edge "bad edge" if it lies inside some $2 \times 2$. Every domino contains exactly one edge and if that edge is bad, then that domino lies inside some square $2 \times 2$(call it bad domino). So we want to minimize the number of bad dominos. Paint all the bad edges. They divide the board into: $4$ of $1 \times 1$ pieces(corner pieces), $196$ of $1 \times 2$ pieces(edge pieces) and some $2 \times 2 pieces$(center pieces) Notice that except for corner pieces each piece's area is even. Starting from a corner piece we will move between pieces by only bad dominos.(We start from a corner piece and the domino that contains that piece is forcefully bad domino. The other part of the domino leads us to another piece. In the next piece before the encounter that piece had even number of squares and after the encounter it now has odd number of squares left. We can say that in the tiling there exist a bad domino that contains only one square of the current piece and that domino leads us to another piece. We can continue this until we reach another corner piece) So we can find a series of bad dominos that reveals a path between to corner pieces. And actually there are two such path since there are $4$ corner piece. Since each corner piece is at least $100$ square apart, each path is at least $50$ domino length. So there are at least $100$ bad dominos. As for example for only $100$ bad dominos, put $100$ vertical dominos on the edges and make other dominos horizontal.

a_507_bc wrote: A square grid $100 \times 100$ is tiled in two ways - only with dominoes and only with squares $2 \times 2$. What is the least number of dominoes that are entirely inside some square $2 \times 2$? Very beautiful. The answer is $100{}$. For the construction, tile the upper and lower rows with horizontal dominoes and the rest of the board with vertical dominoes; clearly, only those on the upper and lower rows lie entirely in some $2\times 2$ square. Consider the segments of the $100\times 100$ board which do not lie on the boundary of any $2\times 2$ square (as illustrated below, in the case of a $6\times 6$ table) and color them red. Note that a domino lies entirely inside a $2\times 2$ square if and only if it contains a red segment inside it. We also color the boundary of the $100\times 100$ board orange. [asy][asy] size(4.5cm); draw((2,0)--(2,6), gray); draw((4,0)--(4,6), gray); draw((0,2)--(6,2), gray); draw((0,4)--(6,4), gray); draw((1,0)--(1,6), red+linewidth(1.1)); draw((3,0)--(3,6), red+linewidth(1.1)); draw((5,0)--(5,6), red+linewidth(1.1)); draw((0,1)--(6,1), red+linewidth(1.1)); draw((0,3)--(6,3), red+linewidth(1.1)); draw((0,5)--(6,5), red+linewidth(1.1)); draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0), orange+linewidth(1.1)); [/asy][/asy] The main observation is the following: if some red and orange segments span a shape of odd area, then that shape cannot be tiled with dominoes, hence a domino must cross its boundary, that is, it contains a red segment inside it (it cannot contain an orange segment inside it, as it would exit the board). Therefore, if we create $100{}$ shapes which have odd areas and disjoint red boundaries, we win! To achieve this, consider all of the $100{}$ squares with odd side-lengths flushed against the upper-left and lower-right corners of the $100\times 100$ board (as illustrated below, in the case of a $6\times 6$ table). This completes the proof. [asy][asy] size(10cm); draw((2-7,0)--(2-7,6), gray); draw((4-7,0)--(4-7,6), gray); draw((0-7,2)--(6-7,2), gray); draw((0-7,4)--(6-7,4), gray); draw((1-7,0)--(1-7,5)--(6-7,5), gray); draw((3-7,0)--(3-7,3)--(6-7,3), gray); draw((5-7,0)--(5-7,1)--(6-7,1), gray); draw((1-7,6)--(1-7,5)--(0-7,5), red+linewidth(1.1)); draw((3-7,6)--(3-7,3)--(0-7,3), red+linewidth(1.1)); draw((5-7,6)--(5-7,1)--(0-7,1), red+linewidth(1.1)); draw((0-7,0)--(0-7,6)--(6-7,6)--(6-7,0)--(0-7,0), orange+linewidth(1.1)); draw((8-(2+1),6-0)--(8-(2+1),6-6), gray); draw((8-(4+1),6-0)--(8-(4+1),6-6), gray); draw((8-(0+1),6-2)--(8-(6+1),6-2), gray); draw((8-(0+1),6-4)--(8-(6+1),6-4), gray); draw((8-(1+1),6-0)--(8-(1+1),6-5)--(8-(6+1),6-5), gray); draw((8-(3+1),6-0)--(8-(3+1),6-3)--(8-(6+1),6-3), gray); draw((8-(5+1),6-0)--(8-(5+1),6-1)--(8-(6+1),6-1), gray); draw((8-(1+1),6-6)--(8-(1+1),6-5)--(8-(0+1),6-5), red+linewidth(1.1)); draw((8-(3+1),6-6)--(8-(3+1),6-3)--(8-(0+1),6-3), red+linewidth(1.1)); draw((8-(5+1),6-6)--(8-(5+1),6-1)--(8-(0+1),6-1), red+linewidth(1.1)); draw((8-(0+1),6-0)--(8-(0+1),6-6)--(8-(6+1),6-6)--(8-(6+1),6-0)--(8-(0+1),6-0), orange+linewidth(1.1)); [/asy][/asy]

$200$, so we should have at least $100$ dominoes inside one of those squares.

The answer is 100, achieved by tiling the left and rightmost columns with vertical dominoes, and the rest of the board with horizontal dominoes. We now prove the bound. To that end, define a cell to be bad if it belongs inside a domino that is entirely inside some one of the $2\times 2$ squares. Note that the four corner squares are bad, as well as a square that is adjacent to each corner (covered by the same domino as the corner). We will now use the following claim to propagate bad cells along the board. Claim: If a square that is not a corner is bad, then there is another cell in a diferent $2\times 2$ square that is also bad. Pf: This proof follows from the geometry of domino coverings. Here is an example: [asy][asy] size(4.5cm); draw((0,0)--(0, 2), gray); draw((0,0)--(0, -2), gray); draw((0,0)--(2, 0), gray); draw((0,0)--(-2, 0), gray); draw((0,0)--(0,-2)--(-1,-2)--(-1,0)--(0,0), red+linewidth(1.1)); draw((0,1)--(0,-1)--(1,-1)--(1,1)--(0,1), blue+linewidth(1.1)+dotted); draw((-1,0)--(1, 0)--(1,1)--(-1,1)--(-1,0), green+linewidth(1.1)+dotted); [/asy][/asy] Above, we are propagating bad squares from the upper square of each red domino; the dotted dominoes cannot co-exist in the same tiling, so at least one of the upper of right square is also bad. $\blacksquare$ It is now clear that there are at least $2\cdot 100$ bad cells, by considering paths of bad dominoes starting at each corner, so at least $100$ dominoes that are entirely contained inside a $2\times 2$ square.

Ru83n05 wrote:

I had the same idea but you can't guarantee that this propagation keeps on going and doesn't hit a wall or itself, or something like that. For example, it's not immediately clear how one would deal with a situation like this, in which case the path from the corner just ends: