1. We first construct the coloring for $k=2n$. Label the cells of the board by $(i, j)$, and color all $\{(i, j) : j = i \text{ or } j = i + 1, i \le n\}$ black, as shown in the figure below:

We now prove that there is only one way to arrange the dominoes. Consider the four regions, upper, lower, left, and right, cut out on the board by the two diagonals.
- First, note that for $i \le n$, the domino covering $(i, i)$ must also cover $(i, i+1)$, that is, it must be placed vertically. This forces all dominoes in the left region to be placed vertically (this can be proven by induction.)
- Similarly, for all $i \le n$, the domino covering $(i, i+1)$ must be placed horizontally, which forces all dominoes in the upper region to be placed horizontally.
- The vertical placement of the dominoes on the left will force the dominoes on the lower side to all be placed horizontally.
- Finally, the dominoes on the right are forced to be placed vertically. In other words, there is only one arrangement.

2. We now prove that $k = 2n$ is the minimum. In other words, suppose $k < 2n$, $k$ cells of the board are colored black, and there exists a covering method $P$; then we want to prove that there exists another covering method $P'$. For convenience, let $D = \{(i, i) : 1 \le i \le 2n\}$ denote the main diagonal of the board.

a. First, construct a graph whose vertices are all the cells of the board, with edges colored red and blue:
– If two cells are covered by the same domino, connect them with a red line;
– If two cells that are symmetric to each other with respect to the main diagonal are covered by the same domino, connect them with a blue line.
Note the following points:
– Two points may be connected by both a red line and a blue line at the same time.
– Every point has degree 1 with respect to red lines and degree 1 with respect to blue lines.
– In other words, we can decompose this graph into several disjoint cycles, each cycle consisting of alternating red and blue lines. Here we accept cycles of length 2, that is, the case where two points are connected by both a red and a blue line.

b. Consider a cell $d \in D$ on the main diagonal. Note that $d$ is symmetric with respect to $D$, so it cannot be connected to the same point by both a red and a blue line at the same time; in other words, it must belong to a cycle of length at least 4, call it $C(d)$.

c. Suppose the vertices of $C(d)$ are $c_0, c_1, \cdots, c_n$, where $c_0 = d$. Let $m$ be the smallest positive integer such that $c_m \in D$; clearly, $c_m \neq d$. But note that by our construction, if we reflect the path connecting $c_0, c_1, \cdots, c_m$ across $D$, the reflected path will also appear in the original graph; in other words, $C(d)$ must consist of $c_0, c_1, \cdots, c_m$ together with its reflection. This means that for every $d \in D$, $C(d)$ has exactly two points belonging to $D$.
In summary, the $2n$ points in $D$ must be distributed among $n$ cycles $C_1, C_2, \cdots, C_n,$ each of which has length at least 4.

d. Now, by the pigeonhole principle, there must be some $C_i$ containing at most one colored black cell (otherwise there would be at least $2n > k$ black cells, a contradiction). We can modify $P$ in the following way to obtain a new covering method $P'$:
- Remove all dominoes corresponding to red lines of $C_i$.
- Place dominoes according to the blue lines in $C_i$.
Since $|C_i| \ge 4$, $P'$ must differ from $P$; and since $C_i$ has at most one black cell, every domino still covers at most one black cell.

This proves that if $k < 2n$, the covering method cannot be unique. Q.E.D.