Romanian Mathematical Olympiad

Suppose, WLOG, that the height of $ABCD$ equals $6$. Denote by $k$ the maximal number of elements of the set $W$ and by $M_1, \dots, M_k$ its elements. For each $i \in \{1, 2, \dots, k\}$ denote by $a_i, b_i, c_i$, and $d_i$ the distances from the point $M_i$ to the planes $(BCD), (ACD), (ABD)$ and $(ABC)$ respectively. Then $a_i, b_i, c_i, d_i \in \{0, 1, 2, \dots, 6\}$. Because the sum of the distances from an interior point to the faces of a regular tetrahedron equals its height, we get that $a_i + b_i + c_i + d_i = 6$. Consequently, if $T = (a_1 + a_2 + \dots + a_k) + (b_1 + b_2 + \dots + b_k) + (c_1 + c_2 + \dots + c_k) + (d_1 + d_2 + \dots + d_k)$, we get $T = 5$.
Let $s = [k/2]$. As no more than two of the numbers $a_1, a_2, \dots, a_k$ can be equal, we deduce $s \le 7$.

For even $k$, $k = 2s$, we have $a_1 + a_2 + \dots + a_k \ge 2 \cdot 0 + 2 \cdot 1 + \dots + 2 \cdot (s-1) = s^2 - s$. So $T = 12s \ge 4s^2 - 4s$, and thus $s \le 4$ and $k \le 8$. For odd $k$, $k = 2s + 1$, we have $a_1 + a_2 + \dots + a_k \ge 2 \cdot 0 + 2 \cdot 1 + \dots + 2 \cdot (s-1) + s = s^2$. Thus $T = 12s + 6 \ge 4s^2$, implying $s \le 3$ and $k \le 7$.

To give an example of an 8-element set, use the notation $(a_i, b_i, c_i, d_i)$ as in the above considerations and consider as $W$ the set $(0, 1, 2, 3), (1, 0, 2, 3), (0, 1, 3, 2), (1, 0, 3, 2), (2, 3, 0, 1), (2, 3, 1, 0), (3, 2, 1, 0)$ and $(3, 2, 0, 1)$.