2. Let $n$ be a positive integer. On a $2 n \times 2 n$ grid, place several dominoes such that each cell of the grid is adjacent to exactly one cell covered by a domino. For each positive integer $n$, find the maximum number of dominoes that can be placed according to the above requirements.

Note: A domino has a size of $1 \times 2$ or $2 \times 1$ grid. Each domino can only be placed exactly on two cells of the grid, and the positions of different dominoes cannot overlap. If two different cells share a common edge, they are called "adjacent".
2. Let $M$ be the maximum number of dominoes that can be placed according to the problem's requirements.
We will prove in two steps that $M=\frac{n(n+1)}{2}$.
(1) Prove by mathematical induction: For any $n \in \mathbf{Z}_{+}$, it is possible to place $\frac{n(n+1)}{2}$ dominoes on the grid satisfying the requirements.
(i) When $n=1,2$, as shown in Figure 1 and Figure 2.
(ii) Assume the conclusion holds for $n=k$, consider the case for $n=k+2$.

Construct the placement based on the parity of $k$ as shown in Figure 3 ($k$ is odd) and Figure 4 ($k$ is even).

In these figures, the middle gray area is placed according to the method for $n=k$ (induction hypothesis), and an additional $2k+3$ dominoes need to be placed in the outermost layer.
Thus, the total number of dominoes is
$$
\frac{k(k+1)}{2} + 2k + 3 = \frac{(k+2)(k+3)}{2}.
$$

Therefore, for $n=k+2$, the conclusion also holds.
Combining (i) and (ii), we conclude that for any $n \in \mathbf{Z}_{+}$, it is possible to place $\frac{n(n+1)}{2}$ dominoes on the grid satisfying the requirements.
(2) Prove that $M \leqslant \frac{n(n+1)}{2}$.
First, extend the $2n \times 2n$ grid to a $(2n+2) \times (2n+2)$ grid (adding $8n + 4$ cells around the perimeter). A cell is considered occupied by a domino if it is covered by a domino or adjacent to a covered cell.

It is easy to see that each domino occupies 8 cells in the extended grid.

Next, consider the occupancy of the $8n + 4$ cells on the outermost layer. Use "曰" to indicate cells around a domino (occupied), "区" to indicate cells not around a domino (unoccupied), and "口" to indicate cells where a domino cannot be placed.

Based on the placement of the dominoes, there are five scenarios as shown in Figure 5.

In these five scenarios, the occupancy ratio of the $8n + 4$ cells around the perimeter, i.e., the ratio of occupied to unoccupied cells, is: $\frac{1}{1}, \frac{1}{2}, \frac{1}{1}, \frac{2}{3}, \frac{3}{7}$.

Thus, at most half of the $8n + 4$ cells around the perimeter can be occupied, i.e., no more than $4n + 2$ cells.
$$
\begin{array}{l}
\text{Therefore, } M \leqslant \left[\frac{4n^2 + 4n + 2}{8}\right] \\
= \left[\frac{n(n+1)}{2} + \frac{1}{4}\right] = \frac{n(n+1)}{2},
\end{array}
$$

where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
Combining (1) and (2), we conclude that the maximum number of dominoes that can be placed according to the problem's requirements is $\frac{n(n+1)}{2}$.