toddm:
snip...2. If i setup my bridge to be balanced at 1000 ohms i will not be able to tell the difference between a positive or negative output of the bridge within my measurement range. A 990 ohm would output about the same voltage as a 1010 ohm but negative.
... snip
Yes, you can tell the difference between positive and negative... and this is the behavior we both want and need.
To expand on what MarkT has said, you want the bridge to be balanced at the value you want to measure. A Wheatstone bridge is nothing more than two independent voltage dividers which uses four resistors with your differential meter measuring between the center points of those two voltage dividers.
To explain in painful detail, lets assume all four resistors (includes the R under test) are exactly 1000 ohms. Also, let's use 10.0 volts dc as the voltage applied to the bridge.
Since all values have been defined, calculate the voltage at the mid-point of each divider circuit using Vo= Vin * r2/(r2+r1) and you'll see it is exactly 0.5 times your applied voltage which gives us five volts at center point since 10.0 * 0.5 = 5.0 volts. Since the bridge is as we say, balanced, there is exactly five volts at the center point of the other two resistors which gives us a differential voltage of exactly zero volts which means our resistor under test is in fact 1000 ohms, no more, no less.
Now, let's say your resistor under test is 1010 ohms. Recalculating the voltage on that side of the bridge yields 0.5025 * Vin giving us +5.025 volts. Since the other side of the bridge has not changed, its output is 0.5 * Vin and the differential voltage is 0.025 volts (5.025-5.0). Since that voltage represents the upper limit of our resistor under test we can use that to set the gain of our amplifier. Again, lets use 10 volts out of our amplifier as the limit so the gain needs to be 10.0/0.025 = 400. So, with an amplifier output of +10.0 volts, the resistor under test is exactly 1% greater than 1000 ohms. +5.0 volts would mean the resistor is 0.5 % high or 1005 ohms.
Let's now flip the test case and say the R under test is 990 ohms. Doing the math gives us 0.4975 for the ratio and plugging that into our amplifier gives an output of 400 * (4.975v - 5.0v) = -10.0 volts. This is same as the example above with the only difference being the sign of the output which means our resistor under test is -1.0% low.
Since you want to do this with an Arduino, you can easily drop the supply voltage to 5 volts and leaving the amp gain at 400 gives you a +/-5 volts amp output equaling a range of +/- 1% tolerance of your 1k resistor under test.
So now hopefully you can see why the center point of the bridge needs to be 1000 ohms?
If you want to increase the test range to say +/- 2%, all that needs to be done is to reduce the differential amp gain to 200 and you're there. Extrapolating to 5% or what ever range you want is just as easy.
As a side topic, the amplifiers used in bridge circuits are usually very high gain with low bandwidths and are "chopper" stabilized to prevent drift from input bias current drift. Have a look at David Jones µCurrent device over at EEVBlog for a quick tour on the current technology in chopper stabilized op amps.