4. (10 points) Remove two numbers from $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}$ so that the sum of the remaining three numbers is closest to $\frac{6}{7}$. The two numbers to be removed are ( )
A. $\frac{1}{2}, \frac{1}{5}$
B. $\frac{1}{2}, \frac{1}{6}$
C. $\frac{1}{3}, \frac{1}{5}$
D. $\frac{1}{3}, \frac{1}{4}$
【Analysis】The sum of these five fractions is 1.45, while $\frac{6}{7} \approx 0.857$, the former is 0.593 greater than the latter, so the problem requires selecting two fractions from the first five such that their sum is closest to 0.593. After comparison, it can be seen that $\frac{1}{3}$ and $\frac{1}{4}$ should be chosen.
【Solution】Solution: $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{5}{6}+\frac{9}{20}+\frac{1}{6}=1 \frac{9}{20}=1.45$,
$$
\begin{array}{l}
\frac{6}{7} \approx 0.857, \\
1.45-0.857=0.593,
\end{array}
$$

So the problem requires selecting two fractions from the five such that their sum is closest to 0.593. After comparison, it can be seen that $\frac{1}{3}$ and
$$
\frac{1}{4} \text {. }
$$

Therefore, the answer is: D.