, 2023

The answer is $1011 + \frac{1}{M}$.

We can see that $C_M \ge 1011 + \frac{1}{M}$ by picking $a_{2i+1} = 1$ and $a_{2i} = -1$. This way, we get for all $k = 1, \dots, M-1$,
$$\left\{ \frac{ka_1}{M} \right\} + \left\{ \frac{ka_2}{M} \right\} + \dots + \left\{ \frac{ka_{2023}}{M} \right\} = 1012 \cdot \frac{k}{M} + 1011 \cdot \frac{M-k}{M} = 1011 + \frac{k}{M} \ge 1011 + \frac{1}{M}.$$

To show that this is the best possible, let
$$P_i := \sum_{k=1}^{M-1} \left\{ \frac{ka_i}{M} \right\}, \quad Q_k = \sum_{i=1}^{2023} \left\{ \frac{ka_i}{M} \right\}.$$
We claim that for any integer $a$, we have $\sum_{k=1}^{M-1} \left\{\frac{ka}{M}\right\} \le \frac{(M-1)}{2}$, and equality holds when $\gcd(a, M) = 1$. To see this, let $d = \gcd(a, M)$. Then clearly
$$\begin{aligned} \sum_{k=1}^{M-1} \left\{ \frac{ka}{M} \right\} & \le (d-1) \cdot \frac{0}{M} + d \cdot \frac{d}{M} + d \cdot \frac{2d}{M} + \dots + d \cdot \frac{(M/d-1)d}{M} \\ & \le \frac{1}{M} \left[ (1 + \dots + (d-1)) + (d + \dots + 2d-1) + \dots + (M-d+\dots+M-1) \right] \\ & = \frac{(M-1)}{2}, \end{aligned}$$
where the equality holds only when $d=1$.

Since $a$ is arbitrary, by applying the claim for $a = a_i$, we see $P_i \le \frac{M-1}{2}$ for all $i$. Thus $\sum_{k=1}^{M-1} Q_k = \sum_{i=1}^{2023} P_i \le \frac{2023(M-1)}{2}$.

Now if there is some $Q_k \le 1011$, then we are already done. Thus we can assume $\lceil Q_k \rceil \ge 1012$ for all $k$, and so we have
$$\{-Q_k\} = -Q_k - \lfloor -Q_k \rfloor = \lceil Q_k \rceil - Q_k \ge 1012 - Q_k.$$
As a consequence,
$$\sum_{k=1}^{M-1} \{-Q_k\} \ge 1012(M-1) - \frac{2023}{2}(M-1) = \frac{1}{2}(M-1).$$

Note that
$$\{-Q_k\} = \left\{ \frac{-k(a_1 + \cdots + a_{2023})}{M} \right\}.$$
Apply the claim for $a = a_1 + \cdots + a_{2023}$, and we get that all equalities should hold. Thus $\gcd(a, M) = 1$ by the claim, and $\lceil Q_k \rceil = 1012$ for all $k$. In particular, if we choose $k$ to be the inverse of $a$ modulo $M$, then $Q_k = 1011 + \frac{1}{M} \le C_M$, as desired.