4.56 Given that $x_{1}, x_{2}, x_{3}, \cdots, x_{1989}, x_{1990}$ are 1990 positive numbers, and
$$
\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\frac{1}{1+x_{3}}+\cdots+\frac{1}{1+x_{1989}}+\frac{1}{1+x_{1990}}=1 \text {. }
$$

Then among the 1990 positive numbers $x_{1}, x_{2}, x_{3}, \cdots, x_{1989}, x_{1990}$,
(A) There is at most 1 number less than 1.
(B) There must be a number greater than 1989.
(C) There must be a number less than 1989.
(D) There must be one that is 1990.
(China Beijing High School Mathematics Competition, 1990)
[Solution] If $x_{1}=x_{2}=\cdots=x_{1990}=x$, then the given expression can be simplified to
$$
\frac{1990}{1+x}=1 \text {, }
$$

Thus, $x=1989$, which rules out options (B), (C), and (D).
Furthermore, the number of $x_{i}$ less than 1 cannot exceed 1, otherwise:
Assume $0\frac{1}{2}+\frac{1}{2}=1 \text {, }
$$

which contradicts the given condition.
Therefore, the correct choice is (A).