Ответ. $(N+1)^2(4N+1)$.

Let $k = (N+1)^2(4N+1)$. Introduce a coordinate system where all vertices of unit cubes have integer coordinates from $0$ to $2N+1$.

First, we present an example showing that the number of black cubes can indeed be $k$. For each cube, consider its vertex closest to the origin (its coordinates range from $0$ to $2N$). Let the cube be black if at least two coordinates of this vertex are even, and white otherwise. Clearly, in any $2 \times 2 \times 2$ cube there will be exactly $4$ black and $4$ white cubes. The number of black cubes with all three corresponding coordinates even is $(N+1)^3$, and the number with exactly two even coordinates is $3(N+1)^2N$, making the total number of black cubes equal to $k$.

Now we prove this example is optimal. Consider a cube partitioned into black and white unit cubes satisfying the problem's conditions. We'll call a cube dark or light if it's respectively black or white in the above example.

For each point $(a, b, c)$ in the large cube, define its $x$-, $y$-, and $z$-rank as $r_x = \min(a, 2N + 1 - a)$, $r_y = \min(b, 2N + 1 - b)$, and $r_z = \min(c, 2N+1-c)$. The rank is $r = \min(r_x, r_y, r_z)$ - the distance to the nearest face of the large cube.

Mark all unit cube vertices with odd ranks. For each marked vertex, consider the difference between the number of black and white cubes meeting at it. Since these vertices are centers of $2 \times 2 \times 2$ cubes, this difference is non-positive, making the total sum $\Sigma$ of such differences non-positive.

Define the multiplicity of a unit cube as the number of its marked vertices. Then $\Sigma$ equals the difference between the sum of multiplicities of black cubes and white cubes. We need to show that if this difference is non-positive, the number $\ell$ of black cubes doesn't exceed $k$.

Let $r_x \le r_y \le r_z$ be the ranks of a cube's center. Then:
- If $r_x < r_y$, its multiplicity is $4$
- If $r_x = r_y = \frac{1}{2} + d$ with even $d$, multiplicity is $< 4$ and it's dark
- If $r_x = r_y = \frac{1}{2} + d$ with odd $d$, multiplicity is $> 4$ and it's light

Thus, dark cubes have multiplicity $\le 4$, light cubes $\ge 4$.

Let $s_1 \le \dots \le s_{(2N+1)^3}$ be the multiplicities in order. From our example where $\Sigma = 0$, we have $s_1 + \dots + s_k - s_{k+1} - \dots - s_{(2N+1)^3} = 0$. If $\ell > k$:
$$\begin{aligned} 0 \ge \Sigma &\ge s_1 + \dots + s_\ell - s_{\ell+1} - \dots - s_{(2N+1)^3} > \\ &> s_1 + \dots + s_k - s_{k+1} - \dots - s_{(2N+1)^3} = 0, \end{aligned}$$
since $s_{k+1} \ge 4$. This contradiction proves $\ell \le k$.