At a guess, the "other" circuits you've seen use U1A as an integrator in the "normal" configuration, with the + input tied to ground.
Your circuit works like this. Assume the output of U1A is positive. Then the output of U1B is positive, and the input to U3A is 0. So the Q* output goes high after the next clock, and the U2A output will go high. This will drive the integrator towards zero. The integrator will then tend to wander around zero, and the circuit will work more or less as a sigma-delta convertor.
However, because you've driven the + input of the integrator, rather than using a standard resistor to -, you've produced a converter which will be distinctly non-linear. This won't be a problem for small input swings and low precision, but it will be a real problem for large inputs. Consider the correction current in the standard configuration - it is independent of the input voltage. Assuming the op amps give +/- 4 volts out when driven into saturation, the correction current into the integrator will be +/- 4V divided by 100K, or 40 uA, regardless of input voltage. In your circuit, a 0 volt input will produce a correction current of 40 uA, but a 2 volt input will produce a correction current of (4-2)/100K, or 20 uA. Like I say, this will probably do OK if the input swings are small, especially if the downstream circuitry only attempts to zero the input, but it will provide inaccurate readings if you're trying to measure the input.