Turning off the LM3914 dot/bar display driver per the datasheet
Thanks to everyone who commented. Based on the responses, I have two solutions for blanking the LM3914. I think the first method gets close to making sense of the datasheet in my original question. The second method illustrates a technique to "gate off" the input signal, also per the datasheet. I tested both of these solutions on the bench and they seem to work fine.

Note: The LM3914 provides 1.25 V via REF_OUT when REF_ADJ is connected to ground. In the following schematics, the internal voltage reference of 1.25 V is tied directly to the divider network, producing 125 mV between steps. I left off the transistor pull-down resistors for the sake of brevity.

Method 1: Shorting the Voltage Reference

The LM3914 display will turn off if no current flows through the voltage reference output REF_OUT. According to the datasheet, "pulling positive on the reference" will turn off the internal current sources, essentially shutting off the chip. In my testing with a 5 V supply voltage, I measured a standby current of only 150 µA in this state.

In the schematic below, R1 sets the LED current according to the datasheet formula \$I_{LED}\approx\frac{12.5}{\mathit{R1}}\$. When Q1 is on, current flows across R1 to ground and the display operates normally. When Q1 is off, REF_OUT is pulled to +5V via R1 and R2, shutting off the display (and just about everything else inside the chip).

Note that if R_HI is tied to REF_OUT, and R_LO is tied to ground (as shown below), leakage through the internal resistor network may keep the display dimly illuminated. Thus, R2 must be sufficiently low to offset the leakage and flow enough current from the +5V rail. I chose 22 kΩ empirically and measured about 180 µA through REF_OUT. If the internal resistor network is connected to a different voltage reference, I've been able to blank the display (and shut off the chip) with as little as 40 µA through REF_OUT. I think the 100 µA mentioned in the datasheet is more of a guideline and depends on the wiring arrangement.



Method 2: Shorting the Input Signal

The LM3914 display will turn off when the input signal is grounded. This can be accomplished with an NPN transistor as shown below. When Q1 is off, the LM3914 sees the signal voltage and operates normally. When Q1 is on, SIG is essentially grounded through the transistor. R2 was chosen to limit the Q1 "on" current to about 125 µA maximum; the input bias current at SIG is only 25-100 nA, so this should be fine. R3 was chosen to ensure Q1 is fully saturated when turned on; it passes about 44 µA and could probably be a lot higher than 100 kΩ. During testing, I found that the \$V_{CE}\$ of Q1 (and thus the voltage on SIG) was less than 30 mV, which is considerably lower than the minimum 125 mV required to illuminate the first LED. So, the display stays off.

Unlike Method 1, which draws hardly any current when blanked, all of the internal current sources are still operating when the input signal is grounded. In my testing with a supply voltage of 5 V, this amounted to about 6 mA. Not much, but still far greater than the 150 µA observed with Method 1.

This approach worked when I built the circuit. Is it an appropriate (i.e., "best practices") method to ground an analog signal?