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 $\lvert trace(H) \rvert \notin \{0,2\sqrt{n},4\sqrt{n},...,n\}$. 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 $n<196$, here 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 $|