8. In a Cartesian coordinate system, if all vertices of a polygon are integer points, then the polygon is called "integral". If an integral convex 14-gon can be divided into integral parallelograms, and the area of each parallelogram does not exceed $C$, find the minimum value that $C$ can take.
8. The minimum value that $C$ can take is 5.

Lemma 1 The given integer-coordinate 14-gon has exactly seven pairs of parallel opposite sides.

Proof of Lemma 1 Since it is a convex 14-gon, it suffices to prove that each side has a parallel opposite side.
For any side $A B$ of this 14-gon, draw a line perpendicular to $A B$ and ensure that all 14 vertices are on the right side of line $A B$. Define the right side of this line as the right direction.
Consider an integer-coordinate parallelogram $P_{1}$, with one side (the left side) being $A B$, and the other side (the right side) on the side of this 14-gon or inside the 14-gon.
If the opposite side is inside the 14-gon, then construct an integer-coordinate parallelogram $P_{2}$, with its left side being the right side of $P_{1}$, and the right side of $P_{2}$ on the side of the 14-gon or inside it. Continue constructing such integer-coordinate parallelograms $P_{1}, P_{2}, \cdots, P_{n}$, until the other side of $P_{n}$ is on a line of one of the sides of the 14-gon, which is parallel to $A B$.
Lemma 1 is proved.
The series of parallelograms $P_{1}, P_{2}, \cdots, P_{n}$ generated by side $A B$ is called the "parallelogram chain" of side $A B$.
Lemma 2 Any two parallelogram chains have a common parallelogram, which we call the intersection of these two parallelogram chains.
Proof of Lemma 2 Consider two pairs of opposite sides $A B / / A^{\prime} B^{\prime}$, $C D / / C^{\prime} D^{\prime}$.
Since the 14-gon is convex, connecting the midpoints of the sides parallel to $A B$ in the parallelogram chain of $A B$ forms a broken line, with the starting and ending points being the midpoints of sides $A B$ and $A^{\prime} B^{\prime}$. This broken line divides the 14-gon into two parts, with sides $C D$ and $C^{\prime} D^{\prime}$ on opposite sides of the broken line.
Thus, consider the broken lines corresponding to the parallelogram chains of $C D$ and $C^{\prime} D^{\prime}$. These two broken lines intersect, indicating that these two parallelogram chains have a common parallelogram.
Lemma 2 is proved.
Select the shorter side of each pair of parallel sides, and among the seven pairs of sides, choose the sides that are not parallel to each other. For each of these sides (denote it as $E F$), let $\boldsymbol{v}(E F)$ be the vector with integer components, of minimum length and parallel to $E F$, called the direction of $E F$, and denote its coordinates as $x(E F)$ and $y(E F)$.
Thus, $\operatorname{gcd}(x(E F), y(E F))=1$.
By Lemma 2, for any two non-parallel sides $A B$ and $C D$, there must exist a parallelogram $P$ whose sides are parallel to $A B$ and $C D$, respectively. Therefore, the area of $P$ can be divided by the area of the parallelogram with sides $\boldsymbol{v}(A B)$ and $\boldsymbol{v}(C D)$.

Finally, it can be proved that there exists a parallelogram that can be divided into five parts.
Let the area of $P$ be $S(P)$. Thus,
$S(P)=|x(A B) y(C D)-x(C D) y(A B)|$.
If $x(A B)$ and $x(C D)$ are both multiples of 5, then $S(P)$ is a multiple of 5.

If no more than one vector's x-coordinate is a multiple of 5, then at least 6 direction vectors have x-coordinates not equal to 0, and
$$
\frac{y(A B)}{x(A B)} \equiv \frac{y(C D)}{x(C D)}(\bmod 5) .
$$

Thus, $S(P)=|x(A B) y(C D)-x(C D) y(A B)|$
$$
=\left|\left(\frac{y(A B)}{x(A B)}-\frac{y(C D)}{x(C D)}\right) x(A B) x(C D)\right|
$$
$\equiv 0(\bmod 5)$.
Since there exists a parallelogram whose area is a multiple of 5, i.e., $C \geqslant 5$, the following example shows that $C=5$ is achievable.
Select the following 14 points:
$$
\begin{array}{l}
(0,0),(1,0),(2,1),(3,3),(3,4), \\
(2,6),(1,7),(-1,8),(-2,8), \\
(-3,7),(-4,5),(-4,4),(-3,2),(-2,1),
\end{array}
$$

This 14-gon can easily be divided into parallelograms with areas not exceeding 5.