As bws48 and chaosmagnet have stated this is not a straight forward question. Its determinate on a variety of factors, from the construction of the motor, duration of the inverter and how much you wish to invest.

Typically, an inverter is a interim power source between loss of power event until the generator is energized. Not certain of how the inverter will be used.

As stated by chaos the modified or clipped sine wave from the inverter will not impact a small motor. The size of the unit is determined by the operation of the pump. Ideally, a pump should run for no less than 5 minutes to prevent short cycling.

Also, you need to consider the depth of the sump. For example, if the sump pit takes 5 minutes to fill and approximately 5 minutes to empty and you require 30+ minutes to return home from work - the inverter will require a capacity to operate 3 maybe 4 cylces of the pump.

Now comes the fun part. What is the construction of the motor? Induction, split case, capacitor, etc.? Each will determine the inrush current required to start the pump moving. From the sound of what you are describing - probably a generic motor requiring about 1000-1100(W) starting and 500-550(W) running. Please note this will not be used in the calculation below, it is just an estimate.

As inverters are rated in VA the load must also be expressed in VA. However, the maximum load applied to an inverter is no more than 75% of its stated capacity (eg. 1000VA cannot exceed 750VA continuous load). To calculate for a given motor:

VA MOTOR RUNNING
[HP * Efficiency (standard motor) * 0.8](kVA) * 1000(VA/kVA) or
HP * 0.746 * 0.8 * 1000 = VA
1/3(hp) * 0.746 * 0.8 * 1000 = 198.93(VA) or 200(VA)

VA MOTOR TOTAL RUN TIME (simplified version of the algorithm)
200(VA) * 20(minutes or 4 cycles) = 4000(VA)

VA MAX LOAD FACTOR ON INVERTER
4000(VA) / 0.75 (max load factor) = 5333(VA)

VA SAFETY FACTOR (aka Cover My Butt)
5333(VA) * 1.5 (safety factor) = 8000(VA)

Hope this helps.