The answer to my question (provided by BS) is the following:

We have to change the action by looking at the group $G=U(n, \mathbb{C})$ and its action by conjugacy on pairs of Hermitian matrices. The space of pairs of such matrices can be identified with $T^* (Lie U(n, \mathbb{C})^*)$ because $Lie U(n, \mathbb{C})$ consists of antihermitian matrices (so we should only divide by $i$ to obtain Hermitian matrices).

The coadjoint orbit is an orbit of a matrix with ones everywhere except for the diagonal (where it has zeroes). Its orbit is all Hermitian matrices $T$ such that $rk (T+\mathrm{Id})=1$. Moment map is $J(X,Y)=-i[X,Y]$. And then a representative of each element in the orbit can be chosen in a form $(X,Y)$ where $X$ is a diagonal matrix and $Y_{j k }= \frac{i}{x_j-x_k}$.

So the idea of AHusain to multiply by $i$ was a good one -- but one has to change the action...

Note that this proof (for $H$) is quite the same as a proof for the potential $H^-$: it is related to the fact that they both come from the group $SL(n, \mathbb{C})$:this group has (among others) two real parts: $SL(n, \mathbb{R})$ and $SU(n, \mathbb{R})$. The first part corresponds to $H^-$ and the second to $H^+=H$.