What is the problem with connecting a negative voltage to an op amp's negative lead while the possitive one is grounded?
Therefore, Vin, which is negative, would be connected to a virtual
ground and, for some reason, this requires to protect the op. amps.
with D1.

First thing, if Vcc is powered down there cannot be a virtual ground so yes, there could be a situation where a negative voltage appears at the inverting input.
Secondly, for every op-amp I've ever used, if \$V_{IN}\$ is fed via a resistor (i.e. \$R_G\$ in the circuit) then it is highly current limited. So, providing that the negative voltage from \$V_{IN}\$ (via \$R_G\$) cannot push more current into the op-amp input pin than what the data sheet tells you is the absolute maximum, then it's a non-event and the diode is not needed.
Typical maximum currents are about 5 mA and, to get 5 mA flowing into the op-amp inverting input pin requires: -

(a) For the pin to be about 0.3 volts below the negative power rail (i.e. starting to cause the chip parasitic diodes to conduct) and...
(b) \$V_{IN}\$ to be -50.3 volts when feeding via the 10 kΩ resistor, \$R_G\$.

Basically, in all but a really small number of applications, you don't need the diode. Here's how much current you can push into an LM324 input pin: -

It's a whopping big 50 mA (an old device) but more modern devices are somewhat less like the AD8605: -

For the MCP6004 it's 2 mA (taken from data sheet): -


If relying on this current as a means of protection, always check the latest data sheet and don't push it to the limit.