The function $\sqrt{t}$ is not defined for negative $t$. So it makes no sense to look at the behaviour as $t$ approaches $0$ through negative values.
The sum, as a function $F(t)$, can be expressed, for $t>0$, in terms of the hyperbolic cotangent of $\pi/\sqrt{t}$. For $t=0$, each term is $0$, so $F(0)=0$.
It was shown by @Shai Covo and @anon that $\lim_{t\to 0+}F(t)=\pi/2$. So $F(t)$ is not continuous (from the right) at $t=0$.
However, if we define the function $G(t)$ by $G(t)=F(t)$ if $t>0$, and $G(0)=\pi/2$, then the function $G$ is continuous (from the right) at $t=0$. This is because $\lim_{t\to 0+}F(t)=\pi/2$.
Of course $G(t)$ cannot be fully continuous at $0$, since we have not even defined it for negative $t$. If we cared to, we could define $H(t)$ by $H(t)=G(t)$ if $t\ge 0$, and $H(t)=G(|t|)$ for $t<0$. Then $H(t)$ would be continuous for all $t$. This is not all that unreasonable. Instead of summing $\sqrt{t}/(1+n^2 t)$, we would be summing $\sqrt{|t|}/(1+n^2 |t|)$.
The discontinuity (from the right) of $F(t)$ at $t=0$ is in a sense not a mathematically significant one. The technical term is that it is a removable discontinuity. The value of $F(0)$ is the "wrong one" for continuity from the right, but that can be easily changed by replacing that value by the "correct one," which should be $\pi/2$.