How to calculate voltage across each resistor in a chain for a given tolerance?
Tolerance in this case refers to the resistor value (R) being accurate. A tolerance of 1% means a 75kΩ resistor could be as low as \$75\times 0.99=74.25\space Ω\$ or as high as \$75\times 1.01=75.75\space Ω\$ (thermal effects aside).

Calculating voltage across a given resistor is an application of Ohm's Law, \$V={I\times R}\$.

If all resistors are the same value, you can take a 'shortcut' to determine voltage across a given resistor (\$V_R\$) by dividing total voltage (\$V_T\$) by the number of resistors (\$R_N\$): \$V_R=V_T/R_N\$. In your example with four resistors, this is \$V_R=1600/4=400\$.

If resistor are not the same value, you will need to apply Ohm's Law to calculate total current first, then find the voltage across each resistor.

$$
R_T=R_1 + R_2 + R_3...\\
I_T=V_T/R_T
$$

When a resistor's actual value is 1% lower than the target value, it will have 1% lower voltage drop across it due to the relationship of \$I\times R\$; remember that current through a resistor (\$I_R\$) will be equal to the total current \$I_T\$, so as \$R\$ varies, so will \$V_R\$.

Note that you can't calculate \$V_R\$ using the shortcut then simply assume the voltage varies as the resistance value varies on a given resistor. Its resistance contributes to the total resistance, which affects total current.

By "worst case" I assume you mean highest voltage, which would occur to a particular resistor if its value was at the higher end of the tolerance range, and all others were at the lower end of the range. If, in your four resistor example, R1 was +1% while all others were -1%, \$R_T\$ would be 298.5kΩ, making \$I_T\$ 5.36mA. This gives the following resistor voltages:


\$V_{R1}\$: 406.0302V
\$V_{R2}\$...\$V_{R4}\$: 397.9899V


I'll leave it to you to expand on this for the 24 resistors as well as greater tolerance values.