Aviov N.I.

$N^{3}$ unit cubes are drilled through the diagonal and tightly strung on a thread, after which the thread is tied into a ring (that is, the vertex of the first cube is connected to the vertex of the last one). For which $N$ can such a necklace of cubes be packed into a cubic box with an edge length of $N$?
Let's choose a cube in the necklace and mark it with the number 1. Then we will number the remaining cubes in sequence, moving along the thread in one of the two possible directions. In the cube with number $n$, we denote by $A_{n}$ the vertex belonging to the thread that adjoins the previous cube.

We choose a coordinate system, aligning the origin with the vertex of the box, directing the axes along its edges, and taking the length of the cube's edge as the unit of length. If the necklace is packed into the box, then the vertices belonging to the thread of each cube have different parities of their x-coordinates. Therefore, the sum of these two x-coordinates for each cube is an odd number. Consequently, in the case of an odd $N$, the sum of all these x-coordinates for all cubes is also an odd number. But each x-coordinate is repeated twice: each vertex $A_{n}$ belongs to two cubes.

Thus, the specified sum is even. Therefore, it is impossible to pack the necklace into the box when $N$ is odd.

When $N$ is even, in each cube, we draw a diagonal connecting a vertex of the form (even, even, odd) (i.e., a vertex where the first two coordinates are even and the third is odd) with a vertex of the form (odd, odd, even). Consider the graph formed by vertices of this type. It is not difficult to understand that this graph is connected. Moreover, each vertex inside the cube is connected to eight vertices, on the face to four, and on the edge to two vertices. Therefore, according to a known criterion (see the solution to problem $\underline{30806}$), there exists a cycle in this graph that passes through each edge exactly once. By arranging the cubes in the order of traversal of this cycle so that the drilled diagonal of each falls on the corresponding edge, we obtain the required arrangement of our necklace.

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