4. Figure 1 is a triangular number table, from top to bottom called the first row, second row, …. It is known that each “ ” in the triangular number table is the reciprocal of a positive integer, and is equal to the sum of the two numbers directly below it. If the number in the first row is $\frac{1}{1}$, then the numbers in the third row from left to right can be filled in ( ).
(A) exactly one way
(B) exactly two ways
(C) exactly three ways
(D) infinitely many ways
4.C.

Let the two numbers in the second row be $m, n$. Then
$$
\frac{1}{m}+\frac{1}{n}=1\left(m, n \in \mathbf{N}_{+}\right) \text {. }
$$

Thus, $m=\frac{n}{n-1}=1+\frac{1}{n-1}$. Solving this, we get $n-1=1$. Therefore, $m=2$, and $n=2$, meaning the numbers in the second row can only be $\frac{1}{2}, \frac{1}{2}$.
Let the two numbers under $\frac{1}{2}$ in the third row be
$$
\frac{1}{2}=\frac{1}{m}+\frac{1}{n}\left(m, n \in \mathbf{N}_{+}\right) \text {. }
$$

Then $m=\frac{2 n}{n-2}=2+\frac{4}{n-2}$.
Hence, $(n-2) \mid 4$, so $n-2=1,2,4$.
Thus, $\left\{\begin{array}{l}m=6, \\ n=3\end{array}\right.$ or $\left\{\begin{array}{l}m=4, \\ n=4\end{array}\right.$ or $\left\{\begin{array}{l}m=3 \text {, } \\ n=6 .\end{array}\right.$
Therefore, the numbers in the third row from left to right are
$\frac{1}{3}, \frac{1}{6}, \frac{1}{3}$ or $\frac{1}{6}, \frac{1}{3}, \frac{1}{6}$ or $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$.