To show it is possible to place $n(n+1)/2$ dominoes, generalize the domino placements below.
To prove this is optimal, given a placement of dominoes, let
- $x$ be the number of interior dominoes,
- $y$ be the number of dominoes with both halves on the border,
- $z$ be the number of dominoes with one half on the border and the other in the interior.
Note that a perimeter domino covers $4$ perimeter squares, and a half-interior domino covers at least three perimeter squares. Since there are $4(2n-1)$ perimeter squares total, we must have
Furthermore, an interior domino covers $8$ squares, a border domino not on the corner covers $6$ squares, an a half-interior domino covers $7$ squares. There are at most $4$ corner dominoes, which each cover one fewer square. Therefore,
Putting this all together, we get
Finally, since $x+y+z$ is an integer, we further conclude that $$x+y+z\le \left\lfloor\frac12n(n+1)+\frac14\right\rfloor=\frac12n(n+1).$$