Problem: Fix a positive integer $n$.
Find the smallest positive integer $\chi$ for which
there exists a tournament on $n$ vertices,
and a coloring of each of the tournament's edges by one of $\chi$ colors, such that:
any two directed edges $u \to v$ and $v \to w$ have different colors.

Solution: The answer is
\[ \chi = \left\lceil \log_2 n \right\rceil. \]

First, we prove by induction on $n$
that $\chi \ge \log_2 n$ 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).
\begin{center}
\begin{asy}
  size(5cm);
  filldraw(box( (-2,-2), (-0.6,2) ), opacity(0.1)+lightcyan, lightblue );
  filldraw(box( (0.6,-2), (2,2) ), opacity(0.1)+lightcyan, lightblue );
  label("$A$", (-2,0), dir(180), blue);
  label("$B$", ( 2,0), dir(  0), blue);
  label("all edges colored $c$", (0,2), dir(90), deepgreen);
  draw( (-1.3,1.2)--(1.5,0.2), deepgreen, EndArrow);
  draw( (-0.9,-0.2)--(1.1, -0.1), deepgreen, EndArrow);
  draw( (-1.4, 0.4)--(0.9, 0.8), deepgreen, EndArrow);
  draw( (-1.4, -0.8)--(1.3, -0.6), deepgreen, EndArrow);
  draw( (-1.3, -0.5)--(1.2, -1.6), deepgreen, EndArrow);
\end{asy}
\end{center}
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
\[ \chi \ge 1 + \log_2 (n/2) = \log_2 n \]
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 $S  =\{0, \dots, n-1\}$.
Instantiate colors $c_1$, $c_2$, \dots.
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 $v \to w$.
Moreover we color the edge with color $c_k$.
This works and uses at most $\left\lceil \log_2 n \right\rceil$ colors.

\begin{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 \emph{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
  $f(n) = 1 + \max \left( f(|A|), f(|B|) \right)$
  and we choose $|A| \approx |B|$.
  This optimization also gives the inductive construction.
\end{remark*}