By an easy argument, I can prove that the sum of the title is $\ll T^{-1/2}$ as $T \to \infty$.
But if we recall the integral version of this sum, then we can prove that \[ \int_{-\infty}^{\infty}(|t - T| + 1)^{-1/2 - h} (|t| + 1)^{-1} dt \ll T^{-1/2}. \] (better improvement on $h$).
Can we generalize the integral case to the sum one?
Or can we find out a better bound for the sum of the title?
Thanks in advance.
A simple argument... Let's assume $T$ is a positive integer, first. Then, we have \begin{align}\sum_{n \in \mathbb{Z}} (|n - T| + 1)^{- 1} (|n| + 1)^{-1}=& \sum_{n<0}(|n - T| + 1)^{- 1} (|n| + 1)^{-1}+\sum^T_{n=0}(|n - T| + 1)^{- 1} (|n| + 1)^{-1}\\&+\sum_{n>T}(|n - T| + 1)^{- 1} (|n| + 1)^{-1} \end{align} Now, we'll estimate those three sums: $$\sum_{n<0}(|n - T| + 1)^{- 1} (|n| + 1)=\sum^\infty_{n=1}\frac1{(n+T+1)(n+1)}=\frac1T\sum^\infty_{n=1}\left(\frac1{n+1}-\frac1{n+T+1}\right),$$ where the RHS is a sort of telescoping sum, so its value is $$\frac1T\sum^{T+1}_{n=2}\frac1n=\frac1T\,(H_{T+1}-1),$$ where $H_n$ are the harmonic numbers.
\begin{align}\sum^T_{n=0}(|n - T| + 1)^{- 1} (|n| + 1)^{-1}=&\sum^T_{n=0}\frac1{(T+1-n)(n+1)}=\frac1{T+2}\sum^T_{n=0}\left(\frac1{n+1}+\frac1{T+1-n}\right)\end{align} Here. the RHS is clearly $$\frac2{T+2}\,H_{T+1}.$$ Finally, $$\sum_{n>T}(|n - T| + 1)^{- 1} (|n| + 1)^{-1}=\sum^\infty_{n=T+1}\frac1{(n-T+1)(n+1)}=\sum^\infty_{n=1}\frac1{(n+T+1)(n+1)}$$ with the translation $n-T\to n$, the same value as the first sum.
In total, we obtain $$\left(\frac2T+\frac2{T+2}\right)\,H_{T+1}-\frac2T,$$ and that's clearly $O\left(\dfrac{\ln T}T\right)$ due to the asymptotic behavior of harmonic numbers. Since it's easy to show that the original sum is monotone decreasing as a function of $T$ (hint: derivative), the same is true for non-integer $T$.