Negative group delay and envelope advance
Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies.

Yes, you are right that it is not uncommon to have practical LTI systems with $+ve$ gradient of $\angle H(f)$ in parts of the response and hence making group delay $-ve$ around those parts of $H(f)$. And, if we can give a Narrow-Band input $x(t)$ such that the bandwidth of $x(t)$ is restricted in that part of $H(f)$, then you would have a time-advanced output. So, are we able to look into the future? 

NO!!!! We are not. My point will get clear in a minute.

Let me take an example of a very common and practically realizable IIR filter in equivalent discrete time scenario : the Leaky Integrator.

The $H(e^{j\omega})$ of a leaky integrator is given by the following:
$$H(e^{j\omega}) = \frac{1-\lambda}{1-\lambda e^{-j\omega}},$$So,
$$|H(e^{j\omega})|^2 = \frac{(1-\lambda)^2}{1 + \lambda^2 -2\lambda cos(\omega)}, \angle{H(e^{j\omega})} = arctan \{\frac{-\lambda sin(\omega)}{1-\lambda cos(\omega)} \}$$

The shape can be plotted in MATLAB by following:


  freqz(0.1, [1 -0.9], (-pi:0.001:pi));




Now, if we can give a very narrow-band input $x[n]$ centered around $\omega = 0.6\pi$ and bandlimited within a very small $\Delta \omega$, then we would get a response as follows:

$x[n] = s[n]cos[\omega_o n]$, where s[n] is a narrowband baseband signal and $\omega_o = 0.6\pi$ and group delay of the filter is $g_d$ around $\omega_o$
$$Y(e^{j\omega}) = X(e^{j\omega}).e^{-j.g_d(\omega-\omega_o)},$$ You can work this out to get $y[n] = s[n - g_d]cos[\omega_o n]$

According to the equation above, leaky integrator is basically producing an output which is having a delayed envelope of input by $g_d$ samples. And, what happens if this $g_d$ is negative!

Check out that $g_d$ is indeed negative around $\omega_o = 0.6\pi$. Does that mean that the leaky integrator is able to produce the $s[n]$ envelope $g_d$ samples in advance? 

No, it is not. The caveat is that we cannot have a perfectly bandlimited narrowband causal input $x[n]$. We cannot have a $x[n]$ which has an absolute start in time and yet it is having a bandlimited narrowband response in frequency domain. 

Because we cannot have such input $x[n]$, hence we cannot have a "future seeing time machine".

In order to produce a causal input, which has an identifiable absolute start in time, the frequency response of the input will spread in frequency domain and the input $X(e^{j\omega})$ will be present at all frequencies with non-zero spectral components, and this will make the overall delay to be positive.

Indeed, if you plot the group delay response of the leaky integrator, you get the following, and check that even though the group delay is small negative number away from $\omega = 0$, it is taking high $+ve$ values around $\omega = 0$:



Hope that answers your question.