I) Let us introduce a collective notation $z_i$, $i\in I$, for OP's $x_i$'s and $y_i$'s (which by the way do not have to be equal in numbers). Here $I$ is a finite index set. We assume that the map $z:I\to \mathbb{C}$ is injective. Also let us introduce a parity $\sigma: I\to \{\pm 1\}$, which is $+1$ for an $x_i$ and $-1$ for a $y_i$. Define a bi-linear skew-symmetric bracket$^1$
$$\tag{1} \{z_i,z_j\}~:=~ \left\{ \begin{array}{ccl} \frac{1}{z_i-z_j} & \text{for} & i\neq j, \\ 0& \text{for} & i= j. \end{array} \right.$$
Then OP's coupled first-order system can be written in Hamiltonian form
$$\tag{2} \mathrm{i}\dot{z}_j
~=~\sum_{i\in I\backslash\{j\}}\frac{\sigma_i}{z_i-z_j}
~\stackrel{(1)+(3)}{=}~\{z_j,H\},$$
with Hamiltonian
$$\tag{3} H~:=~ -\sum_{i\in I} \sigma_i z_i. $$
More generally, for a function $f=f(z)$, the time evolution is given as
$$\tag{4} \mathrm{i}\frac{d f}{dt}
~=~\{f,H\}+ \mathrm{i}\frac{\partial f}{\partial t}. $$
II) The corresponding second-order system is the Calogero-Moser equations$^2$
$$\tag{5} -\ddot{z}_j
~=~\sum_{i\in I\backslash\{j\}}\frac{1+\sigma_i\sigma_j}{(z_i-z_j)^3}. $$
It is a major point that the sum on the rhs. of eq. (5) only runs over elements of the same kind, i.e. if ${z}_j$ on the lhs. is an $x_j$, then the non-zero terms in the sum on the rhs. is only over the $x_i$'s, i.e. independent of the $y_i$'s. Hence the evolution of $x_j$ only depend on the $y_i$'s via their initial conditions [1]. And vice-versa with the roles $x_i \leftrightarrow y_i$ exchanged.
III) The Calogero-Moser Hamiltonian in Darboux coordinates $(z_i,p_i)$ reads
$$ \tag{6}H_{CM}~=~\frac{1}{2}\sum_{i\in I}p_i^2
+\frac{1}{4}\sum_{i,j\in I}^{i\neq j}\frac{1+\sigma_i\sigma_j}{(z_i-z_j)^2}. $$
IV) In the spirit of the Kazhdan-Kostant-Sternberg (KKS) construction [2,3], let us define a position matrix
$$ \tag{7} Z~:=~\text{diag}(z_i),$$
a parity matrix
$$ \tag{8} \Sigma~:=~\text{diag}(\sigma_i), \qquad \Sigma^2 = {\bf 1},$$
and a momentum matrix
$$ \tag{9} P_{ij}~:=~\left\{ \begin{array}{ccl}
\frac{\mathrm{i}}{z_j-z_i} & \text{for} & i\neq j, \\
p_i& \text{for} & i= j. \end{array} \right.$$
The diagonal elements $p_i$ of the momentum matrix (9) have a physical interpretation as initial velocities. The momentum matrix (9) satisfies the following Canonical Commutation Relation (CCR) for finite-dimensional matrices:
$$ \tag{10} [Z,P]-\mathrm{i}{\bf 1}~=~\text{rank-one matrix}.$$
The Calogero-Moser Hamiltonian (6) can then be written as
$$ \tag{11} H_{CM}~=~\frac{1}{4}{\rm Tr}(P^2+P\Sigma P\Sigma).$$
The flow of $z_i(t)$ is given by the eigenvalues of the matrix
$$\tag{12} Z(t=0)+t\frac{P(t=0)+\Sigma P(t=0)\Sigma}{2}. $$
The matrix (12) is block diagonal consisting of two blocks. Each block is just the standard KKS construction. In particular, the eigenvalues $x_i(t)$ and $y_i(t)$ do only talk to each other via the initial conditions.
References:
M. Stone, I. Anduaga, and L. Xing, The classical hydrodynamics of the Calogero–Sutherland model, J. Phys. A: Math. Theor. 41 (2008) 275401, arXiv:0803.3735.
D. Kazhdan, B. Kostant, and S. Sternberg, Hamiltonian group actions and dynamical systems of Calogero type, Comm. Pure Appl. Math. 31 (1978) 481.
P. Etingof, Lectures on Calogero-Moser systems, arXiv:math/0606233.
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$^1$ Note that the bracket (1) does not satisfy the Jacobi identity. We suspect that the bracket (1) can be extended to a homotopy hierarchy of higher brackets, although we did not pursuit the matter, partly because the bracket (1) is not important for the rest of the answer.
$^2$ To prove Eq. (5) from eq. (2), the following identity is helpful:
$$\mathrm{i}(\dot{z}_i -\dot{z}_j) ~=~\{z_i-z_j,H\}$$
$$\tag{13} ~\stackrel{(1)+(3)}{=}~-\frac{\sigma_i+\sigma_j}{z_i-z_j}
+(z_i-z_j)\sum_{k\in I\backslash\{i,j\}}\frac{\sigma_k}{(z_i-z_k)(z_j-z_k)}.$$
