2. Given a positive integer $m$, and $n \geqslant m$, in an $m \times 2 n$ grid, what is the maximum number of dominoes $(1 \times 2$ or $2 \times 1$ small grids) that can be placed satisfying the following conditions:
(1) Each domino exactly covers two adjacent small grids;
(2) Any small grid is covered by at most one domino;
(3) Any two dominoes cannot form a $2 \times 2$ grid;
(4) The last row of grids is exactly covered by $n$ dominoes.
2. The maximum value sought is $m n-\left[\frac{m}{2}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$), and for any alternating two rows, using $n$ pieces of $1 \times 2$ and $n-1$ pieces of $1 \times 2$ dominoes respectively ensures that the squares of the last row are exactly covered by $n$ dominoes.

To prove that the number of dominoes satisfying the conditions of the problem is at most $m n-\left[\frac{m}{2}\right]$, label the rows of the original grid from bottom to top as the 0th, 1st, ..., $(m-1)$th rows, and draw $n-i$ fictitious domino blocks symmetrically on the $i$-th row (thus, the $i$-th row has $i$ small squares on each side that are not included). Figure 2 shows the case where $m=n=6$.

A fictitious domino block is called "good" if it is exactly covered by one domino; otherwise, it is called "bad."
We will discuss the following cases.
(1) If all fictitious domino blocks are good, then the remaining dominoes will not cover these fictitious domino blocks. Therefore, these dominoes must be distributed in the triangular regions of side length $m-1$ at the upper left and upper right corners.

Color the small squares like a chessboard, and we know that the number of black squares and white squares in any triangular region differs by $\left[\frac{m}{2}\right]$.

Since each domino covers two different colored small squares, there must be at least $\left[\frac{m}{2}\right]$ small squares in each triangular region that are not covered, hence the conclusion holds.
(2) To handle the remaining case where there are bad fictitious domino blocks, we need to prove that a covering method satisfying the conditions can be transformed into another, ensuring that the number of dominoes does not decrease, and the number of bad fictitious domino blocks decreases. After a finite number of transformations, all domino blocks can be made good, and the number of dominoes does not decrease, so the conclusion holds by case (1).

As shown in Figure 3, consider the row with the smallest label that contains a bad domino block (this is clearly not the last row) and let $D$ be this domino block. Let $l$ and $r$ be the small squares to the left and right of $D$, respectively. Note that the small squares below $l$ and $r$ are covered by $D_{1}$ and $D_{2}$, respectively.

If $l$ is covered by a domino $D_{l}$, then since $D$ is bad and no two dominoes can form a $2 \times 2$ square, $D_{l}$ must be vertical. If $r$ is also covered by a domino $D_{r}$, then $D_{r}$ is also vertical and forms a $2 \times 2$ square with $D_{l}$, leading to a contradiction. Therefore, $r$ must be empty. Thus, the domino $D_{l}$ can be replaced with $D$, still satisfying the problem's conditions, and ensuring that the number of dominoes does not decrease, while the number of bad fictitious domino blocks decreases.
The case where $r$ is covered by a domino is similar.
If both $l$ and $r$ of $D$ are not covered by dominoes, a domino can be directly added at $D$; or to ensure that no two dominoes form a $2 \times 2$ square, the domino directly above $D$ can be moved to the position of $D$.