Let $A = [a_{ij}]_{n\times n}$ be a Hermitian matrix, such that $|a_{ij}| =1$ for $i \neq j$, and $a_{ii} = 0$ for each $i$. I am interested in a tight lower bound of $\|A\|_*:=\sum_{i=1}^n |\lambda_i(A)|$, where $\lambda_i(A)$'s are eigenvalues of $A$.

Note that, by minimizing $\sum_{i=1}^n |\lambda_i|$ over two constraints $\sum_{i=1}^n \lambda_i = 0$ and $\sum_{i=1}^n \lambda_i^2= n(n-1)$, one can obtain $\sqrt{2n(n-1)}$ as a lower bound. But it seems that isn't tight.

On the other hand, if $A := J - I$ (all ones matrix minus identity), then $\sum_i |\lambda_i(A)| = 2(n-1)$.

Is it true that $2(n-1)$ is actually a lower bound (for large enough matrices, say $n \geq 10$) ?

**Remarks:**

As Alex's answer below, the minimum of trace norm of such matrices may be less than $2(n-1)$, even for arbitrarily large matrices.

But, as a comment of @fedja, the minimum is $(2+o(1))n$ as $n\to\infty$.

**Added:**

- In the particular case, when $a_{ij}=±1$, the lower bound holds. See this answer below, for an overview of the proof.