$16 \cdot 16$ On a straight line, there are four distinct points in sequence $A, B, C, D$. The point with the minimum sum of distances to $A, B, C, D$
(A) can be a point outside the line $A D$.
(B) is only point $B$ or $C$.
(C) is only the midpoint of segment $A D$.
(D) has infinitely many points.
(China Junior High School Mathematics League, 1987)
[Solution 1] It is easy to prove that the sum of the distances from any point on line segment $BC$ to the four points $A, B, C, D$ is equal to $AD + BC$, which is also the minimum of the sum of the distances from any point in the plane to $A, B, C, D$.
Therefore, the answer is $(D)$.
[Solution 2] The shortest distance between two points on a plane is a line segment, so the point we are looking for should be on $AD$.
Let the point we are looking for be $X$ and $A, B, C, D$ correspond to $x, a, b, c, d$ on the number line, respectively, and it is clear that $a < b < c < d$. (As shown in the figure).
\begin{tabular}{ccccc}
$a$ & $b$ & $x$ & $c$ & $d$ \\
\hline$A$ & $B$ & $X$ & $C$ & $D$
\end{tabular}

Thus, the problem is reduced to finding the minimum value of
$$
y = |x - a| + |x - b| + |x - c| + |x - d|
$$

Note that: if $a_{1} < a_{2} < \cdots < a_{n}$, then
$$
f(x) = |x - a_{1}| + |x - a_{2}| + \cdots + |x - a_{n}| \text{ ( } x \text{ is a real number) }
$$

The minimum value is \quad\left\{\begin{array}{ll}f\left(a_{(n+1) / 2}\right), & n \text{ is odd; } \\ f(t), & n \text{ is even; }\end{array}\right.
At this time \quad $a_{n / 2} \leqslant t \leqslant a_{(n / 2)+1}$.
Therefore, $X$ should be on $BC$.
Hence, the answer is $(D)$.