1. Understanding the Problem:
We are given a figure $A$ formed by marked cells on a checkered plane. The figure $A$ satisfies two conditions:
- Any cell of the figure $A$ has exactly two adjacent cells of $A$.
- The figure $A$ can be divided into isosceles trapezoids of area $2$ with vertices at the grid nodes, and the acute angles of the trapezoids are $45^\circ$.

2. Analyzing the Isosceles Trapezoids:
Each isosceles trapezoid has an area of $2$. Given that the acute angles are $45^\circ$, the trapezoids are right-angled isosceles trapezoids. The height of each trapezoid is $1$ unit, and the bases are $1$ unit and $2$ units respectively.

3. Midlines of Trapezoids:
Consider the midlines of each trapezoid. The midline of a trapezoid with bases $1$ and $2$ units is $1.5$ units. However, for simplicity, we can scale down the problem such that each midline is $1$ unit.

4. Properties of the Polygon Formed by Midlines:
- Every side of the polygon is either vertical or horizontal and has an integer length.
- The sides alternate between vertical and horizontal.
- If a side has an even length, the adjacent sides go in opposite directions (neighboring angles sum to $360^\circ$). If a side has an odd length, the adjacent sides go in the same direction.

5. Proving the Perimeter is Divisible by 4:
We need to show that the perimeter of the polygon formed by the midlines is divisible by $4$.

6. Carving Out Squares:
- If there are two adjacent sides with lengths greater than $2$, we can carve out or attach a $2 \times 2$ square to decrease the sum of the squares of all sides while preserving the perimeter.
- Eventually, we arrive at a polygon where out of any two adjacent sides, one has a length $\leq 2$.

7. Handling Sides of Length 1:
- If there are two adjacent sides of length $1$, the polygon is a $1 \times 1$ square with a perimeter of $4$.
- If there is a side of length $1$, we can cut out a $1 \times 2$ square from the side of length $1$, preserving the perimeter modulo $4$.

8. Final Case:
- If all side lengths are even, we can scale down by $\frac{1}{2}$, and any lattice polygon has an even perimeter.
- If there is an odd side, we keep carving out $2 \times 2$ squares until the odd side has only a length $1$ part left. Then, we use the previous operation on sides of length $1$ to strictly decrease the perimeter.

9. Conclusion:
- Eventually, we must arrive at a $1 \times 1$ square, and we have preserved the perimeter modulo $4$ with each operation.
- Therefore, the number of marked cells is divisible by $8$.

$\blacksquare$