An Olympic sequence is a sequence $s_{1}, s_{2}, \ldots, s_{2023}$ where each of the 2023 terms is equal to 1 or -1. A slowly increasing sequence is a sequence of integers $t_{1}, t_{2}, \ldots, t_{n}$ such that $1 \leqslant t_{1}<t_{2}<\cdots<t_{n} \leqslant 2023$, and each of the $n-1$ differences $t_{i+1}-t_{i}$ is 1 or 2.
Find the largest integer $C$ such that, for any Olympic sequence $s_{1}, s_{2}, \ldots, s_{2023}$, there exists an integer $n \geqslant 1$ and a slowly increasing sequence $t_{1}, t_{2}, \ldots, t_{n}$ of length $n$ such that

$$
\left|s_{t_{1}}+s_{t_{2}}+\cdots+s_{t_{n}}\right| \geqslant C
$$
A study of small cases leads us to consider the sequence $\left(s_{i}\right)$ whose terms are $+1,+1,-1,-1,+1,+1,-1,-1, \ldots$, and which seems to be a "worst" case. It is divided into 1011 blocks of length 2, denoted $B_{1}, B_{2}, \ldots, B_{1011}$, and a block $B_{1012}$ of length 1: the sequence is $(-1)^{i+1}$ on each block $B_{i}$.
We now consider a sum of the form $s_{t_{1}}+s_{t_{2}}+\cdots+s_{t_{n}}$. Its terms $s_{t_{i}}$ have been chosen from consecutive blocks $B_{u}, B_{u+1}, \ldots, B_{u+v-1}$. At most $(v+1) / 2$ of these 1 blocks contain terms +1; the others contain terms -1. Among the numbers $s_{t_{1}}, s_{t_{2}}, \ldots, s_{t_{n}}$, there are therefore at most $v+1$ that are +1 and at least $(v-1) / 2$ that are -1. Since $v \leqslant 1012$, we deduce that $s_{t_{1}}+s_{t_{2}}+\cdots+s_{t_{n}} \leqslant(v+1)-(v-1) / 2=(v+3) / 2-508$, so that $C \leqslant 507$.
Conversely, let $(s_{i})$ be any Olympic sequence. By replacing each term with its opposite if necessary, we assume that $(s_{i})$ has at least 1012 terms equal to +1. As before, we subdivide this sequence into maximal blocks on which $(s_{i})$ is constant; these blocks are denoted $B_{1}, B_{2}, \ldots, B_{k}$, and their lengths are denoted $b_{1}, b_{2}, \ldots, b_{k}$. Assuming $B_{1}$ is empty if necessary, we assume that $s_{j}=(-1)^{i}$ when $j$ belongs to a block $B_{i}$. Thus, $b_{2}+b_{4}+b_{6}+\cdots \geqslant 1012$ and $1011 \geqslant b_{1}+b_{3}+b_{5}+\cdots$
We then choose $\left(t_{i}\right)$ as the non-increasing sequence that takes all values $j$ belonging to a block $B_{2}, B_{4}, B_{6}, \ldots$ or located in an even position of a block $B_{1}, B_{3}, B_{5}, \ldots$ In this way, among the numbers $s_{t_{1}}, s_{t_{2}}, \ldots$, there are $b_{2}+b_{4}+b_{6}+\cdots$ that are +1; each of $\left\lfloor b_{1} / 2\right\rfloor+\left\lfloor b_{3} / 2\right\rfloor+\left\lfloor b_{5} / 2\right\rfloor+\cdots$ other terms is -1. We conclude that

$$
\begin{aligned}
s_{t_{1}}+s_{t_{2}}+\cdots & =\left(b_{2}+b_{4}+b_{6}+\cdots\right)-\left(\left\lfloor b_{1} / 2\right\rfloor+\left\lfloor b_{3} / 2\right\rfloor+\left\lfloor b_{5} / 2\right\rfloor+\cdots\right) \\
& \geqslant\left(b_{2}+b_{4}+b_{6}+\cdots\right)-\left(b_{1}+b_{3}+b_{5}+\cdots\right) / 2 \\
& \geqslant 1012-1011 / 2>506
\end{aligned}
$$

so that $s_{t_{1}}+s_{t_{2}}+\cdots \geqslant 507$.
In conclusion, the integer $C$ sought is $C=507$.
Examiner's Comment This problem was generally well done. However, many students rushed into things, neglecting to rigorously prove why, in the case of a sequence $1,1,-1,-1,1,1,-1,-1, \ldots$, it was impossible to have a sum greater than 507 (in absolute value). Generally, any use of a phrase such as "it is clear that" immediately suggests to the examiner that, contrary to what the student claims, "he does not see at all how to prove that."
Moreover, many students rushed into things in their handling of edge cases, thus proposing answers equal to 506 or 508 rather than 507. It is a pity to lose a point in this way when one has essentially finished the problem.