smc

Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes. Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$. The midpoint of this diagonal is at $\left(\frac{3}{2},\frac{3}{2},\frac{3}{2}\right)$. The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac{9}{2}$. The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z<\frac{9}{2}<(x+1)+(y+1)+(z+1)=(x+y+z)+3$. Therefore the cube is intersected by this plane if and only if $x+y+z\in \{2,3,4\}$. There are six cubes such that $x+y+z=2$: permutations of $(1,1,0)$ and $(2,0,0)$. Symmetrically, there are six cubes such that $x+y+z=4$. Finally, there are seven cubes such that $x+y+z=3$: permutations of $(2,1,0)$ and the central cube $(1,1,1)$. That gives a total of $\boxed{19}$ intersected cubes. Note that there are only 8 cubes that are not intersected by our plane: 4 in each of the two opposite corners that were connected by the original diagonal.