When using a potential divider to lower a voltage, think in terms of the upper resistor dropping the bulk of the voltage, leaving the lower resistor with the desired voltage left across it.
It's the ratio of the resistors and not the absolute values that is important.
In equation terms the voltage across the lower resistor
Vl = Rl x Vin/(Ru +Rl)
Where Ru and Rl are the values upper and lower resistors
Consider your 14V example. Lets say that we want to use the upper resistor to drop 10V, leaving 4V across the lower resistor - which is an ideal voltage for a digital input.
The resistors will drop a voltage proportional to their value, so how about a 10K to drop the 10V and a 3.9K (nearest common value to 4K) to drop the remainder?
To get a more formal calculation for how this works - you have to use Ohm's Law.
Back to your 14V example. First think of the resistors added together in series to get the total resistance i.e. 14K.
Then the current flowing through each resistor is the total voltage (14V) divided by the sum of the resistors (14K) so a current of 1mA flows through each of them. Once you know this, you can multiply this current by the resistance if the lower resistor (4K) which will tell you that it will drop 4V and the 10K will drop 10V.
Common handy ratios:
To half the voltage use resistors of equal value - say 10K and 10K
To divide by 4, make the bottom resistor say 1K, and the total = 4K, so the top resistor is 3K.
To divide by 10, the sum has to add up to 10, so make the upper 9K and the lower 1K.
As others have stated - you can ignore the leakage current drawn by the input port.
However you do have to bear in mind the power dissipated in the resistor. Many small resistors will be "quarter watt" so have a maximum power handling of 250mW. They will get hot if you get close to that power rating.
If you drop 10V across 10K the power dissipated will be 10mW. If you drop 10V across 1K the power is 100mW.
Hope the above helps.