Currently, this is an open problem. So I only got a partial answer.
We just need to consider the following two different situations:
Case 1: $n$ is not a perfect square. There is no solution if $\lvert trace(H) \rvert \neq 0$; and $0$ is a possible choice for $\lvert trace(H) \rvert$ if and only if symmetric HM conjecture is correct.
Case 2: $n$ is a perfect square. There is no solution if $\lvert trace(H) \rvert \notin \{0,2\sqrt{n},4\sqrt{n},...,n\}$; and all possible choices are known. So we only need to check $0,2\sqrt{n},4\sqrt{n},...,n$ one by one and make sure at least one instance exists. It is completely settled for small cases$n<196$, for examplehere is my result:
| $n$ | All possible choices for $\lvert trace(H) \rvert$ |
|---|---|
| $1$ | $ 1 $ |
| $4$ | $ 0,4 $ |
| $16$ | $ 0,8,16 $ |
| $36$ | $ 0,12,24,36 $ |
| $64$ | $ 0,16,32,48,64 $ |
| $100$ | $ 0,20,40,60,80,100 $ |
| $144$ | $ 0,24,48,72,96,120,144 $ |