Let the length of the escalator be $L$.

When the escalator is stationary:

• The boy walks up in time $t_1$.
• So his walking speed is $v_b = \frac{L}{t_1}.$

When the boy stands still on the moving escalator:

• The escalator alone takes him up in time $t_2$.
• So the escalator's speed is $v_e = \frac{L}{t_2}.$

When the boy walks on the moving escalator:

• His speed relative to ground becomes $v_b + v_e = \frac{L}{t_1} + \frac{L}{t_2}.$

If the required time is $t$, then $L = \left(\frac{L}{t_1} + \frac{L}{t_2}\right)t.$

Cancel $L$ from both sides: $1 = \left(\frac{1}{t_1} + \frac{1}{t_2}\right)t.$

Therefore, $t = \frac{1}{\frac{1}{t_1} + \frac{1}{t_2}} = \frac{t_1 t_2}{t_1 + t_2}.$

Match with options: $\boxed{\frac{t_1 t_2}{t_1 + t_2}}$ which is Option C.

Comparison with stored answer:

• Stored correct answer: C
• Derived answer: C
• Hence, they agree.