What's a good sensor for measuring high voltages
This answer was written when the question only mentioned 600 V. Since then the 600 V \$\pm\$ 100 V requirement was added.

All you need is a microcontroller with an ADC (Analog-to-Digital Converter) and a resistive voltage divider. Let's presume you'll use a 5 V microcontroller.  



VIN is your solar strings' 600 V, VOUT is the 5 V to the microcontroller's ADC. The 600 V will cause a current of I = 600 V / (R1 + R2), and that same current causes a voltage I &times; R2 across R2, so that  

\$V_{OUT} = \dfrac{R2}{R1+R2} V_{IN}\$

Let's look at the required current first, and then get our resistor values from that. 
Nick is right when he mentions the 1 &micro;A leakage of the microcontroller's I/O pin, but I don't agree with his measures. Having 10 times the leakage cuurent through your voltage divider gives you a 10% error at 600 V, when you easily can go for 1% by choosing a 100 &micro;A current.  



The blue graph shows the maximum error in % due to the 1 µA leakage current, as a function of input voltage, at 100 &micro;A through the divider. You can see that it stays below 2% for voltages higher than 300V. The purple curve shows the error if you use 10 times high resistor values, like Nick does. So 100 &micro;A is OK, 10 &micro;A is too little.

600 V &times; 100 &micro;A = 60 mW, so we probably won't have problems there. 600 V/100 &micro;A = 6 M&Omega;, so a 5.95 M&Omega; + 50 k&Omega; divider will give you 5 V out at 600 V in. For the 5.95 M&Omega; you can use 1% Yageo HHV-25 resistors, a 4.75 M&Omega; and a 1.2 M&Omega; in series give you the required 5.95 M&Omega;. The HHV-25 series has a working voltage of 1600 V and a 1/4 W power rating. For the 50 k&Omega; any 1% metal film resistor will do.