$2 \cdot 36$ The equation $x^{2}+1993 x+3991=0$ has the following root conditions:
(A) It has positive roots.
(B) It has integer roots.
(C) The sum of the reciprocals of the roots is less than -1.
(D) None of the above.
(Huanggang Region, Hubei Province, China Junior High School Mathematics Competition, 1993)
[Solution] Given that the roots of the equation are $x_{1}, x_{2}$, by Vieta's formulas,
$$
x_{1}+x_{2}=-1993 \text {, and } x_{1} x_{2}=3991 \text {. }
$$

We know the equation has no positive roots, so option $(A)$ is false.
Also, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{-1993}{3991}>-1$, so option $(C)$ is false.
If the equation has integer roots, then from $x_{1} x_{2}=3991$ we know the last digits of $x_{1}, x_{2}$ can only be 1,1; 3,7 or 9,9.

Thus, the last digit of $x_{1}+x_{2}$ can only be an even number (0,2,4,6,8), which contradicts $x_{1}+x_{2}=$ -1993. Therefore, option $(B)$ is false.
Hence, the correct choice is $(D)$.