Why is this resistor connected in such way?
If your Opamp have negative power supply tied to the ground, you want to keep signal on it's inputs not lower, than GND and not higher than Vcc. That is what your circuit does - it introduces bias to AC sine wave to keep it in its range. Circuit, you drawn will probably reduce it's gain, because instead of ground R2 is connected to voltage divider R3-R4(which have too high output impedance) and your AC source is also connected to that divider through C2-R1. So part of your AC signal will transform to common-mode signal(because of R1) on Opamp inputs. In other words all that circuitry helps you to "emulate" Opamp powered with negative supply, and R3/R4 divider introduces some "virtual ground" (not to be confused with negative input in inverting amplifier). You have very high gain and with R2 connected to the ground you would have Opamp output always at high rail voltage (VCC - 0.7 or something about that), because you will amplify your signal on positive input with respect to ground, having Vcc/2 offset on your positive input. If you want to connect R2 to ground and keep gain than high, you should chose resistors in your voltage divider very carefully with respect to AC amplitude, which is slightly harder.

To reduce that bias circuitry influence on your circuit gain you may want to: 


Use lower values of R3 and R4 resistors. It is better to have them 10 or more times lower, than R1.
Use separate voltage divider for R2. That may lead to some problems, connected with resistance tolerance of the real parts.


BTW you better check voltage on your C2 cap - maybe it should be replaced with non-polarized.