Maximum Current Input for 555
With any electronic circuit, you make current available to it and the circuit draws it under the pressure of the applied voltage. The circuit allows a certain flow.

Your 555 requires a certain current to operate its internals. It can also deliver a current from its V+ supply pin to its OUTPUT pin. And it can conduct a current coming in on its OUTPUT pin down to its GND pin.

There are various terms for the current to just operate a circuit without its outputs connected to anything (usually). 'Operating current', 'supply current' and 'quiescent current' are some of the most common. The circuit needs this current at a minimum. The datasheet may state min(imum), typ(ical) and max(imum) values for it. Unless the values are significantly high or your power supply is significantly limited, your should allow for the specified maximum current in your power supply design and routing. But you can expect it to normally draw the specified typical current.

The OUTPUT pin driver is a totem-pole output, consisting of a transistor between V+ and OUTPUT and another transistor between OUTPUT and GND. Each can conduct up to 200 mA reliably, although the voltage dropped and lost across each transistor increases as the current through increases and so does the power dissipated in the device. If that power becomes too high, it can damage the chip or heat it up and affect its behaviour to some degree. It is good practice to only draw the minimum you need from the output of a device for the reliable operation of your load.

So if you connect a 1 K resistor between your 555's OUTPUT and GND and run it from a 15 V supply, you can expect the current drawn by the 555 through V+ to be around:

\$I_{total} = I_{q} + I_{load} = 10 mA + (15 - 0.2)/1000 = 24.8 mA\$

where \$I_{q}\$ is the quiescent (supply) current and \$I_{load}\$ is the V+ voltage minus the drop across the output transistor (roughly and from the datasheet) divided by the load resistance. I actually worked the load current out twice:  once to get an idea of the load current then again as I now could introduce the OUTPUT drop at that load (0.2 V) into the calculation.

Going with these these values, you'd expect only the 10 mA to come out of the 555's GND pin because the 14.8 mA is going through OUTPUT and the 1 K resistor to the GND supply rail. 

If you connect a 1 K resistor between your 555's OUTPUT and V+ and run it from a 15 V supply, you can expect the current drawn by the 555's V+ pin to be around:

\$I_{total} = I_{q} = 10 mA\$

This is because the 1 K resistor draws its current from the V+ supply rail separately, not through the 555's V+ pin. But all of the supply current Iq and the load current will flow down and out through the 555's GND pin to the ground rail.