Fix a positive integer $n$. A tournament on $n$ vertices has all its edges colored by $\chi$ colors, so that any two directed edges $u \rightarrow v$ and $v \rightarrow w$ have different colors. Over all possible tournaments on $n$ vertices, determine the minimum possible value of $\chi$.
The answer is
$$
\chi=\left\lceil\log _{2} n\right\rceil .
$$

First, we prove by induction on $n$ that $\chi \geq \log _{2} n$ for any coloring and any tournament. The base case $n=1$ is obvious. Now given any tournament, consider any used color $c$. Then it should be possible to divide the tournament into two subsets $A$ and $B$ such that all $c$-colored edges point from $A$ to $B$ (for example by letting $A$ be all vertices which are the starting point of a $c$-edge).
One of $A$ and $B$ has size at least $n / 2$, say $A$. Since $A$ has no $c$ edges, and uses at least $\log _{2}|A|$ colors other than $c$, we get
$$
\chi \geq 1+\log _{2}(n / 2)=\log _{2} n
$$
completing the induction.
One can read the construction off from the argument above, but here is a concrete description. For each integer $n$, consider the tournament whose vertices are the binary representations of $S=\{0, \ldots, n-1\}$. Instantiate colors $c_{1}, c_{2}, \ldots$. Then for $v, w \in S$, we look at the smallest order bit for which they differ; say the $k$ th one. If $v$ has a zero in the $k$ th bit, and $w$ has a one in the $k$ th bit, we draw $v \rightarrow w$. Moreover we color the edge with color $c_{k}$. This works and uses at most $\left\lceil\log _{2} n\right\rceil$ colors.
Remark (Motivation). The philosophy "combinatorial optimization" applies here. The idea is given any color $c$, we can find sets $A$ and $B$ such that all $c$-edges point $A$ to $B$. Once you realize this, the next insight is to realize that you may as well color all the edges from $A$ to $B$ by $c$; after all, this doesn't hurt the condition and makes your life easier. Hence, if $f$ is the answer, we have already a proof that $f(n)=1+\max (f(|A|), f(|B|))$ and we choose $|A| \approx|B|$. This optimization also gives the inductive construction.