Isn't (as was originally asked) the closest psd matrix to $H=U^* \Delta U$ where $U$ is unitary and $\Delta$ real diagonal with diagonal entries $d_i$ given by $G=U^* \Delta_+ U$ where $\Delta_+$ is the diagonal matrix with diagonal entries $d_i'= \max(0,d_i)$? Because $X\mapsto U^* X U$ is an isometry with respect to the Hilbert distance?
Added after the question was changed. If the closest psd matrix with trace 1 to $H$ is sought, chose $\Delta_+$ to be closest to $\Delta$ subject to the trace 1 and non-negativity condition. The $U$ stuff above reduces your problem to a $\mathbb R^n$ problem from an $n\times n$ matrix problem. I think this is attained when $d_i'$ of the form $d_i'=0$ if $d_i<0$ and $d_i' = d_i+\lambda$ for $d_i\ge0$, where $\lambda$ is chosen to make $\sum d_i' = 1.$