What should be the voltage divider resisor values in units?
Figure 1. The LM358 input bias current parameter.

The input bias current tells you the current that will be supplied by the input pins. (See Figure 16 to understand that the chip will source current due to the PNP input transistors.) It is 100 nA. (I cropped a couple of columns from the data table so the units are missing.)

As with all designs you have a trade-off to consider for your voltage divider. Lower value resistors will make the bias current less of a problem. Higher value resistors will waste less power and this might be critical for a battery-powered application.

If faced with the problem I would consider the following two options:


Set the voltage divider current to 100 times that of the bias current so that the divider should be within about 1% of calculated value. That would be 10 µA in this case and for a 5 V supply would give a total divider chain resistance of \$ R = \frac {V}{I} = \frac {5}{10µ} = 0.5 \ \text M\Omega \$. Any values less that this would improve the situation.
Assume the typical value of 45 µA and correct the divider to allow for this.