Circuit resistances with activating switches
The entire circuit would probably look something like this:







simulate this circuit &ndash; Schematic created using CircuitLab 

The values I computed above came from this formula:

$$R_{{\mid\mid}_N}=3.96\:\text{k}\Omega\left[\frac{R_N}{3.96\:\text{k}\Omega-R_N}\right]$$

where \$R_N\$ is the desired total resistance that you provided in your question and where \$R_{{\mid\mid}_N}\$ is the computed value to use in the circuit, as shown above.

I've used exact values, which in general are not available exactly as shown. But they can be fabricated out of pairs of resistors, if need be.

Now take a look at this chart (borrowed from here):



This lists all of the possible 2-digit and 3-digit resistor values you can get. The E3 are \$\pm 40\%\$, E6 are \$\pm 20\%\$, E12 are \$\pm 10\%\$, E24 are \$\pm 5\%\$, E48 are \$\pm 2\%\$, E96 are \$\pm 1\%\$, and E192 are \$\pm 0.5\%\$.

So, depending on how much accuracy you require, you may have to hunt around to find the right combinations. You can select two resistors from the list and place them in series to get the value you want.

For example, \$3.96\:\text{k}\Omega\$ can't be found in the above table. The closest you can find will be \$3.97\:\text{k}\Omega\$ in the E192 list. But E192 parts are more expensive, too. Regardless, one option is that you could simply accept the value you can get as an E192. It's probably close enough.

However, you might use \$3.96\:\text{k}\Omega=1.47\:\text{k}\Omega+2.49\:\text{k}\Omega\$, where those two resistor values on the right side can be found as either E96 parts (1%) or E48 parts (2%), which are much more common and less costly. Of course, the tolerance is also wider, too.

The same goes for \$R_1\$, \$R_2\$, and \$R_3\$. If you can't find an exact match in the above table and if you want to go to the extra effort to get as close as possible, you can compose those values using pairs of other values to get there. More resistors. But they are cheap.

Of course, you may not need to be anywhere near that precise for your use. Only you know, for sure. But those are some ways to go.

Finally, I think you can see that pressing \$SW_4\$ will indeed cause \$0\:\Omega\$. It shorts out \$R_5\$ so that the resistance between \$T_1\$ and \$T_2\$ will be \$0\:\Omega\$ when \$SW_4\$ is pressed.