8. Given that $n$ is a positive integer. Find the smallest positive integer $k$ such that in a $2 n \times 2 n$ grid, marking $k$ cells ensures there is a unique way to tile the $2 n \times 2 n$ grid with $1 \times 2$ and $2 \times 1$ dominoes, and no domino contains two marked cells.
8. The smallest positive integer $k=2n$.

First, construct an example of marking $2n$ cells that satisfies the conditions: Let the rows and columns be denoted as the 1st, 2nd, ..., 2n-th rows and the 1st, 2nd, ..., 2n-th columns, respectively, and the cell at the intersection of the i-th row and j-th column be denoted as $(i, j)$.

For $i=1,2, \cdots, n$, mark the cells $(i, i)$ and $(i, i+1)$, as shown in Figure 7.
Next, prove that the partition satisfying the conditions is unique.
Notice that the two diagonals of the grid divide the grid into four regions.

Also notice that the domino covering cell $(1,1)$ must be vertical. Therefore, the dominoes covering cells $(2,2), (3,3), \cdots, (n, n)$ are also vertical. By recursive reasoning, all dominoes in the left region are vertical. By rotational symmetry, all dominoes in the bottom region are horizontal; all dominoes in the right region are vertical; and all dominoes in the top region are horizontal, as shown in Figure 8.

If only $k(k<2n)$ cells are marked, and there exists a partition $P$ that satisfies the conditions, we will prove that there exists another partition different from $P$ that also satisfies the conditions.

Let $d$ be the main diagonal of the grid. Construct a graph where each edge is colored one of two colors: each vertex in the graph corresponds to a cell in the grid. If two cells belong to the same domino in $P$, then a red edge is drawn between the corresponding vertices in the graph. A blue edge is drawn between the symmetric points of the two vertices about $d$. It is possible for two vertices to be connected by edges of both colors. Thus, each vertex has a degree of 1 for both red and blue edges. Therefore, the graph is partitioned into several cycles, each of which alternates between red and blue edges (a cycle can have a length of 2).

Consider any cell $c$ on the diagonal $d$. The two edges adjacent to $c$ are symmetric about $d$. Therefore, the cells adjacent to $c$ are different. This indicates that the cycle $C(c)$ containing $c$ has a length of at least 4. Consider a part of this cycle: $c_{0}, c_{1}, \cdots, c_{m}$, where $c_{0}=c$ and $m$ is the smallest positive integer such that cell $c_{m}$ is on the diagonal $d$. Then cell $c_{m}$ is different from $c$. Since the symmetric path about the diagonal $d$ is also part of this cycle, these two parts together form $C(c)$. Therefore, $C(c)$ contains exactly two cells on the diagonal $d$. This indicates that all $2n$ cells on the diagonal $d$ belong to $n$ cycles $C_{1}, C_{2}, \cdots, C_{n}$, each of which has a length of at least 4.

By the pigeonhole principle, there exists a cycle $C_{i}$ that contains at most one of the $k$ marked cells.

To modify $P$, first remove the dominoes corresponding to the red edges in $C_{i}$, and then place the dominoes corresponding to the blue edges in $C_{i}$. Since $C_{i}$ contains at least 4 vertices, the partition $P^{\prime}$ is different from $P$, and since $C_{i}$ contains at most one marked cell, no domino in $P^{\prime}$ contains two marked cells. This proves that the partition is not unique.
Therefore, $k$ cannot be less than $2n$.