Well, when we think about protective diodes we should remember about ESD
(electro static discharges).  Since the primordial of this 3 letters,
everyone heard about protective diodes connecting all pins to Vcc, is
that right?
This could be.  Most of the devices produced with that intention had
diodes anodes to Vcc, but all of them? I am not so sure, "diodes" here
could be a die composition of a transistor. What happens if you apply a
signal 10Volts bigger than Vcc to any I/O pin?  You will be raising VCC
through those diodes, right?  What happens if you apply 500V for 30us at
any I/O pin?  ESD can reach much higher voltages, y'all know that.
Suppose your circuit just have a high impedance alkaline battery (3
Ohms) driven directly the PIC, with just one 10uF capacitor as a Vcc
filter... you can calculate how much the Vcc will raise with that
500V/30us.

I will not say it happens, but it is possible that to avoid that, some
companies use a voltage clipping circuit at each I/O pin, referenced to
VCC (or Ground) as clipping point.  It means that then input clipping
circuit is in true deviating to [Ground] the *internal* input +voltage,
or to Vcc then *internal* input -voltage, the one that would be driven
to the circuit, not exactly what is present at the pin, via somehow
20-100 Ohms input serial (high level) impedances... easy to do.  So, by
this way in presence of Vin > Vcc, Vcc does not rise, or Vin < Ground,
Ground does not goes negative, just an internal chip function clipping
it to ground or Vcc.

Think about as a Protective Gas Capsules or Solid State devices to
protect phone lines and power lines... it doesn't conduct high energy
surge back to the line, doesn't make sense, so it deviates it to ground.

I think it is easy to verify.  Just attach a PIC to a stable +5V Vcc via
a 200 Ohms resistor. If it consumes 5mA, voltage at PIC Vcc pin would be
+4Vdc. Now connect the same supply +5Vcc via another 200 Ohms resistor
to any I/O pin, and measure dc voltage at that I/O pin and also at PIC
Vcc pin, both referenced to ground.  If the PIC Vcc pin increases
voltage from 4V to 4.2V than an internal protective silicon diode (0.6V
drop) connecting I/O pin to Vcc is present.

This should be the "common" diagram:

         Measures 4.0Vcc at this point
                            |
               200 Ohms     V
+5Vcc----o-----RRRRR--------o---PicVccPin (5mA)
              1V drop       |
                            |     NoVdrop
                            '------|<------.
                                           |
                            ----PicI/OPin--o---


         Measures 4.2Vcc at this point
                            |
               200 Ohms     V
+5Vcc----o-----RRRRR--------o---PicVccPin (5mA)
         |     0.8Vdrop     |
         |     4mA          |     0.6Vdrop
         |                  '------|<------.
         |     1mA                         |
         +-----RRRRR------------PicI/OPin--o---
               200 Ohms
               0.2Vdrop

The above consider that the PIC consume still 5mA for 4V or 4.2Vdc.
Is there any good soul that can check if my calculations are correct?