High voltage (70V) amplifier

This is a kind of silly question, but not in a bad way, it can be used as a teachable moment.

The issue you have isn't so much that resistors can be poorly matched, it is that you're assuming op-amps are ideal. That is, "the only source of error is from the resistors, so if I get rid of those I will reach nirvana".

Even if you get matched resistor made from un-obtainium you'll not achieve your (unstated - or rather unparameterized) goal.

In reality all op-amps have issues with input current balancing, input voltage offsets, frequency dependant behaviour etc. Your solution must account for all of these.

And the parameters of your solution space will dictate what can be ignored and what can't be corrected. Hint -> this means you need to give more details as to what is needed, BW, etc.

One classical technique is to use a chopper stabilized amplifier which switches back and forth and self corrects for op-amp and in some topologies, even the resistor network non-linearity.

Another technique is to use switched cap techniques.

Perhaps I'm missing something in your requirements, but something as simple as this would work for DC voltages depending on the current levels needed:

^{simulate this circuit – Schematic created using CircuitLab}

Granted there would be a small voltage drop across the diode, but if you used a power Schottky diode that would only be ~500mV for high current loads, but could be as low as ~100mV for low current signals.

Basically if \$V_1 > V_2\$, then \$D2\$ will be reverse biased and \$D1\$ will conduct making \$V_{out} \approx V_1\$. The opposite will be true if \$V_2 > V_1\$ which would make \$V_{out} \approx V_2\$.

It is worth being aware that you need to pay attention to the reverse leakage current of the diodes you are using. Diodes are not ideal, and will conduct a small current under reverse bias. If the source driving your input is very high impedance compared to the reverse leakage of the diode, you will find that a voltage will appear at the input as the current is driven back through the diode.

You can work out how much this voltage will be by treating the reverse biased diode as a current source based on the reverse current specification from the diode datasheet.