Diagonalizing the positive semi-definite matrix $A$ (say $n\times n$) and assuming $0<l<L<\infty$, we reduce the problem to the following:

What are the best upper and lower bound on $\sum_1^n \frac1{t_j}$ over all positive $t_j$'s such that $l\le\sum_1^n t_j\le L$?$

If $n\ge2$, we can take $t_1=(L+l)/2$ and $t_2,\dots,t_n$ close to $0$. So, we see that there is no finite upper bound on $\sum_1^n \frac1{t_j}$.

On the other hand, we have the inequalities $$\sum_1^n \frac1{t_j}\ge n^2\Big/\sum_1^n t_j\ge\frac{n^2}L,$$ and the lower bound $\frac{n^2}L$ on $\sum_1^n \frac1{t_j}$ is exact, as it is attained when $t_j=L/n$ for all $j$.

Diagonalizing the positive semi-definite matrix $A$ (say $n\times n$) and assuming $0<l<L<\infty$, we reduce the problem to the following:

What are the best upper and lower bounds on $\sum_1^n \frac1{t_j}$ over all positive $t_j$'s such that $l\le\sum_1^n t_j\le L$?$

If $n\ge2$, we can take $t_1=(L+l)/2$ and $t_2,\dots,t_n$ close to $0$. So, we see that there is no finite upper bound on $\sum_1^n \frac1{t_j}$.

On the other hand, we have the inequalities $$\sum_1^n \frac1{t_j}\ge n^2\Big/\sum_1^n t_j\ge\frac{n^2}L, \tag{1}\label{1}$$ and the lower bound $\frac{n^2}L$ on $\sum_1^n \frac1{t_j}$ is exact, as it is attained when $t_j=L/n$ for all $j$; the first inequality in \eqref{1} follows immediately from (say) the HM-GM-AM-QM inequalities.