Answer. For $n=77$.

Solution. Attach four cells to the right and below each domino so that they form a $2 \times 3$ rectangle (if the domino is vertical) or a $3 \times 2$ rectangle (if it is horizontal). If the rectangles of two dominoes have at least one common cell, then the dominoes have a common point. Therefore, if 1014 dominoes, which do not have common points, fit into an $n \times n$ square, then all the constructed rectangles of 6 cells each must fit into a $(n+1) \times (n+1)$ square, obtained by adding a row below and a column to the right of the $n \times n$ square. Hence,

$$
(n+1)^{2} \geqslant 6 \cdot 1014=6084=78^{2}
$$

Thus, $n \geqslant 77$.

It remains to provide an example where this is possible for $n=77$. For this, place the dominoes horizontally in the first row, starting from the first cell, with a gap of one cell; a total of 26 dominoes will fit. Do the same with all other odd-numbered rows. The total number of dominoes will then be $26 \cdot 39=1014$; they obviously do not have common points.
5. In an $n \times n$ square, there are 1014 dominoes (each covering two adjacent cells). No two dominoes share any points (even corner points). For what smallest $n$ is this possible?