For a complete graph with $n \geq 3$ vertices, we want to color its edges using $k$ colors such that an edge can have multiple colors, but

(1) no single color appears on every edge,

(2) no single edge has all the colors, and

(3) in any triangle, every color appears on an odd number of edges.

What is the minimum value of $k$?
Consider an $n$-vertex graph that has been colored with $k$ colors in accordance with the conditions. We assign to each vertex of the graph a sequence of length $k$ consisting of zeros and ones. Let $A$ be a vertex of the graph. We assign to it the sequence consisting entirely of zeros. For any other vertex $B$ different from $A$, the $i$-th element of the sequence assigned to $B$ is 0 if the $i$-th color appears on the edge $AB$, and 1 if it does not.

If $B$ and $C$ are two vertices different from $A$ and from each other, and the $i$-th element of the sequences assigned to them is the same, then the $i$-th color either appears on both edges $AB$ and $AC$, or on neither; since in the triangle $ABC$ the (3) condition requires it to appear an odd number of times, it must appear on the edge $BC$. If, however, the $i$-th elements of the sequences assigned to $B$ and $C$ are different, then the $i$-th color appears on exactly one of the edges $AB$ and $AC$, so it does not appear on $BC$.

In summary, from the (3) condition, it follows that the $i$-th color appears on any edge $XY$ if and only if the $i$-th element of the sequences assigned to them is the same. The converse of this is also true: if the $i$-th color appears on any edge $XY$ if and only if the $i$-th element of the sequences assigned to them is the same, then in any triangle, the $i$-th element of the sequences assigned to the vertices is either the same (in which case the $i$-th color appears on all three edges), or exactly two are the same (in which case the $i$-th color appears on exactly one edge of the triangle).

Condition (2) is equivalent to not assigning the same number to two vertices.

Condition (1) means that there are no two vertices to which the sequences assigned are complements of each other (i.e., the 0s are replaced by 1s and vice versa).

The conditions are therefore equivalent to the fact that from each of the following pairs of sequences

$$
\begin{array}{rr}
(0,0,0, \ldots, 0,0), & (1,1, \ldots, 1,1) \\
(0,0,0, \ldots, 0,1), & (1,1, \ldots, 1,0) \\
(0,0,0, \ldots, 1,0), & (1,1, \ldots, 0,1) ; \\
\vdots & \\
(0,1,1, \ldots, 1,1), & (1,0, \ldots, 0,0) ;
\end{array}
$$

at most one appears on the vertices. Since the number of pairs of sequences is $2^{k-1}$, it follows that $2^{k-1} \geq n$.

Conversely, if $2^{k-1} \geq n$, then the necessary $n$ (length $k$) number sequences can be selected, and the appropriate coloring can be derived from them. The necessary and sufficient condition for colorability is therefore that $2^{k-1} \geq n$, i.e., $k \geq\left\lceil\log _{2} n\right\rceil+1$. Based on the works of Tamás Kiss and Gábor Lippner (Fazekas M. Főv. Gyak. Gimn., III. o.t.).