5. Let $k, n$ be integers, $k \geqslant 2, k \leqslant n \leqslant 2 k-1$. Some $1 \times k$ or $k \times 1$ rectangular pieces of paper are placed on an $n \times n$ grid, such that each piece exactly covers $k$ cells, and no two pieces overlap, until no more pieces can be placed. For any such $k, n$, find the minimum possible number of pieces placed in all configurations that meet the requirements.

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5. When $n=k$, the required minimum value is $n$;

When $k<n<2k$, the required minimum value is $\min \{n, 2n-2k+2\}$.
When $n=k$, the conclusion is obvious.
Assume $k<n<2k$ below.
If $k<n<2k-1$, then
$$
\min \{n, 2n-2k+2\}=2n-2k+2 \text{. }
$$

At this time, $2n-2k+2$ pieces of paper can be placed in the grid as follows.

First, place four pieces of paper in the square area $[0, k+1] \times[0, k+1]$, covering the regions
$$
\begin{array}{l}
{[0, k] \times[0,1],[0,1] \times[1, k+1],} \\
{[1, k+1] \times[k, k+1],[k, k+1] \times[0, k] .}
\end{array}
$$

Then, place $n-k-1$ pieces of paper horizontally in the area $[1, k+1] \times[k+1, n]$, and place $n-k-1$ pieces of paper vertically in the area $[k+1, n] \times[1, k+1]$. Thus, a total of $2n-2k+2$ pieces of paper are placed, and it is easy to see that no more pieces can be placed.
If $n=2k-1$, then
$$
\min \{n, 2n-2k+2\}=n=2k-1 \text{. }
$$

At this time, $n=2k-1$ pieces of paper can be placed in the grid as follows.

First, place $k-1$ pieces of paper horizontally in the rectangular area $[0, k] \times[0, k-1]$, and place $k-1$ pieces of paper horizontally in the rectangular area $[0, k] \times[k, 2k-1]$. Then, place one piece of paper covering the area $[k-1,2k-1] \times[k-1, k]$. Thus, a total of $2k-1$ pieces of paper are placed, and it is easy to see that no more pieces can be placed.

Next, we prove that in any valid placement method, at least $\min \{n, 2n-2k+2\}$ pieces of paper are placed.

Let the number of rows and columns in the valid placement method that do not contain a complete piece of paper be $r$ and $c$, respectively.
If $r=0$ or $c=0$, then at least $n$ pieces of paper are placed.
If $r$ and $c$ are both not equal to 0, we will prove that at least $2n-2k+2$ pieces of paper are placed in this case.

In a valid placement method, consider a horizontal piece of paper $T$.

Since $n<2k$, the nearest horizontal boundary line of the grid to the row where $T$ is located is at most $k-1$ rows away. These rows intersect with the $k$ columns that $T$ crosses, forming a rectangular area where no vertical pieces of paper can be placed. Thus, each of these rows must contain a horizontal piece of paper. Therefore, all such $r$ rows and $n$ columns form an $r \times c$ rectangular area, and it is easy to see that
$$
r<k, c<k .
$$

Thus, there are at least $n-r \geqslant n-k+1$ rows, each containing at least one horizontal piece of paper, and at least $n-c \geqslant n-k+1$ columns, each containing at least one vertical piece of paper. Therefore, at least $2n-2k+2$ pieces of paper are placed.
In summary, the required minimum value is
$$
\min \{n, 2n-2k+2\}(k<n<2k) .
$$