Problem 1208

1. Understanding the Problem:
We are given a cube with edge length $2n + 1$. This cube is to be dissected into smaller cubes of size $1 \times 1 \times 1$ and bars of size $2 \times 2 \times 1$. We need to find the least possible number of $1 \times 1 \times 1$ cubes in such a dissection.

2. Initial Observation:
The total volume of the cube is $(2n + 1)^3$. Each $1 \times 1 \times 1$ cube has a volume of 1, and each $2 \times 2 \times 1$ bar has a volume of 4. We need to cover the entire volume using these pieces.

3. Minimum Number of Small Cubes:
We need to show that at least $2n + 1$ small cubes are required. Consider a plane of size $1 \times (2n + 1) \times (2n + 1)$ within the cube. This plane cannot be completely covered by the $2 \times 2 \times 1$ bars alone because each bar covers an even number of unit cubes (4), and the plane has an odd number of unit cubes $(2n + 1)^2$. Therefore, at least one $1 \times 1 \times 1$ cube is required for each such plane.

4. Inductive Construction:
We will use induction to show that the cube can be filled with exactly $2n + 1$ small cubes.

- Base Case:
For $n = 0$, the cube has edge length $2 \cdot 0 + 1 = 1$. It is already a $1 \times 1 \times 1$ cube, so we need exactly one small cube. This matches $2 \cdot 0 + 1 = 1$.

- Inductive Step:
Assume that a cube of edge length $2k + 1$ can be filled with $2k + 1$ small cubes and the rest with bars. We need to show that a cube of edge length $2(k + 1) + 1 = 2k + 3$ can be filled with $2(k + 1) + 1 = 2k + 3$ small cubes.

Consider the $2k + 1$ cube in the center of the $2k + 3$ cube. By the inductive hypothesis, this can be filled with $2k + 1$ small cubes. The remaining volume forms a shell around this central cube. This shell has a thickness of 1 unit and can be divided into six faces, each of size $2k + 3 \times 2k + 3 \times 1$.

For each face, we can place $2k + 1$ bars of size $2 \times 2 \times 1$ such that only two opposing corners are left empty. This leaves two small cubes per face, totaling $2 \times 6 = 12$ small cubes. However, these 12 small cubes are shared among the faces, and the total number of unique small cubes required is $2k + 3$.

5. Conclusion:
By induction, the least number of $1 \times 1 \times 1$ cubes required to fill a cube of edge length $2n + 1$ is $2n + 1$.

The final answer is $\boxed{2n + 1}$.