High-side current sensing diagram
I guess I can't avoid the temptation. Try:







simulate this circuit &ndash; Schematic created using CircuitLab

In both examples above, you want to use an ultra low current supply, rail-to-rail i/o opamp, with low offset voltage and current. I think the LT1494 may be at least interesting here. The opamp needs to be in control with the inputs near the top rail (the bottom rail won't matter much.)

The opamp supply's current will be just equal to the current in \$R_2\$ (plus its own stray requirements, which is why you want a very low supply current opamp), sinking it through \$R_3\$. Since the input node voltages are approximately equal to each other, this means the current in \$R_2\$ is \$I_{LOAD}\cdot \frac{R_1}{R_2}\$. So the left side's voltage across \$R_3\$ is just a proportional output where \$V_{prop}\approx I_{LOAD}\cdot\frac{R_1\cdot R_3}{R_2}\$. The exact value depends. But in this case I don't think precision isn't terribly important to you.

The right side uses this developed voltage to drive a BJT indicator output. This output is the exact opposite of what you say you want. But I'm sure you can work out how to invert it.

[I selected \$R_5\$'s value to be pretty high (mostly because I'm thinking about keeping the added circuit's current load fairly low and I didn't want to add much load to the voltage developed across \$R_3\$.)]

I've set this to generate an active output when the load current rises over about \$10\:\textrm{mA}\$. You specified zero, but you know that is impossible here. Some reasonable value has to be used. I picked this. You can pick something else, if you want.



Of course, the dedicated ICs (such as those mentioned by Ali Chen) are probably a better way to go.