Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers with absolute values less than 1. What is the minimum value of $n$ if

$$
\left|x_{1}\right|+\left|x_{2}\right|+\ldots+\left|x_{n}\right|=1989+\left|x_{1}+x_{2}+\ldots+x_{n}\right|
$$
According to the condition, each number on the left side is less than 1, so the value on the left side is less than $n$. Thus,

$$
n>1989+\left|x_{1}+\ldots+x_{n}\right| \geqq 1989
$$

which means $n$ is at least 1990. This is possible, for example, if every term on the left side is equal to $\frac{1989}{1990}$, and on the right side,

$$
\left|x_{1}+x_{2}+\ldots+x_{1990}\right|=0
$$

Both conditions are satisfied if 995 of the 1990 numbers are $\frac{1989}{1990}$, and the other 995 are $-\frac{1989}{1990}$.

Of course, for any $n>1990$, there exist real numbers $x_{1}, x_{2}, \ldots, x_{n}$ with absolute values less than 1 such that

$$
\left|x_{1}\right|+\left|x_{2}\right|+\ldots+\left|x_{n}\right|=1989+\left|x_{1}+x_{2}+\ldots+x_{n}\right|
$$

is satisfied: for example, 1990 of them can be chosen according to the above solution, and the rest can be zero.