A cube C of side $n > 2, n ∈ N,$ is divided in $n^3$ cubes of side 1, with disjoint interiors two by two.
We say that two of the cubes of side 1 are Olympic, if any plane parallel to any of the faces of cube C intersects at most one of the interiors of these cubes. We choose the cubes $C_1, C_2,…, C_n$ olympics two by two,and denote with $O_1,…, O_n$ their centers. Determine the minimum value of the sum $O_1O_2+…+O_{n-1}O_n$ and establish which are the configurations formed by n cubes of side 1 for which this minimum is reached.
I think the minimum is $(n-1)\sqrt 3$ but I don’t know to prove.
Following my hint in the comments, let's solve this problem of $n \times n$ unit squares.
We replace planes by lines. In an $n \times n$ grid, if any line parallel to the gridlines does not intersect the interior of more than one square, all squares must belong on different rows and columns. (This is equivalent to placing $n$ mutually non-attacking rooks on an $n \times n$ chessboard.)
Now, we show $O_iO_{i+1} \ge \sqrt 2$.
By Pythagoras, $O_{i}O_{i+1} = \sqrt{d_c^2 + d_r^2}$, where $d_c$ is the difference in there column numbers, and $d_r$ is the difference in their row numbers. By the hypothesis, $d_c,d_r \ge 1$. So, $$O_{i}O_{i+1} \ge \sqrt{1^2+1^2} = \sqrt 2$$
Analogously, in the 3D version, show that $O_{i}O_{i+1} \ge \sqrt 3$, by noting that $$O_{i}O_{i+1} = \sqrt{d_x^2+d_y^2+d_z^2}$$ where $d_x$ is the difference in their $x$ co-ordinates, etc.
Since you already have a construction for the equality case, the claim is proven.