Voltage divider resistance/wattage sizing
You can use a voltage divider made of two resistors so that your voltage sensor can safely read the voltage.

Here are its design equations:

$$
\begin{split}
&amp;\frac{V_\mathrm{in}} {Z1 + Z2} \cdot Z_2 = V_\mathrm{out}\\
\\
&amp;\frac{(V_\mathrm{in} - V_\mathrm{out})^2}{Z_1} =\text{ Power dissipated by }Z_1\\
\\
&amp;\frac{{V_\mathrm{out}}^2}{Z_2} =\text{ Power dissipated by }Z_2
\end{split}
$$

If \$V_\mathrm{out} &lt; \frac{V_\mathrm{in}}{2}\$ I suggest choosing \$Z_1\$ first since it will need to dissipate more power than \$Z_2\$.

Else choose \$Z_2\$ first it will need to dissipate more power than \$Z_1\$.



Edit:

Since the input impedance of your voltage sensor isn't \$\infty\$, there will be a voltage drop at \$V_\mathrm{out}\$. To minimize this voltage drop, you need to use the lowest \$Z_1\$ and \$Z_2\$ value possible. The lower these values get, the more power \$Z_1\$ and \$Z_2\$ need to dissipate. You have to make a compromise between power consumption / dissipation and precision. One thing is for sure, \$Z_1 = Z_2\$ because you need \$V_\mathrm{out}\$ to be \$\frac{V_\mathrm{in}}{2}\$. Here are the minimum \$Z_1\$ and \$Z_2\$ value if you are using \$\frac{1}{4}W\$, \$\frac{1}{2}W\$ or \$1W\$ resistor:

\$\frac{1}{4}W\$ resistor:

\$Z_1 = Z_2 = \frac{{V_\mathrm{out}}^2}{Power} = \frac{{250V}^2}{0.25W} = 250k\Omega\$

\$\frac{1}{2}W\$ resistor:

\$Z_1 = Z_2 = \frac{{V_\mathrm{out}}^2}{Power} = \frac{{250V}^2}{0.5W} = 125k\Omega\$

\$1W\$ resistor:

\$Z_1 = Z_2 = \frac{{V_\mathrm{out}}^2}{Power} = \frac{{250V}^2}{1W} = 62.5k\Omega\$