$4 \cdot 57$ Given that $a_{1}, a_{2}, a_{3}, \cdots, a_{1989}, a_{1990}$ are all positive numbers. Also,
$$
\begin{array}{l}
M=\left(a_{1}+a_{2}+a_{3}+\cdots+a_{1989}\right)\left(a_{2}+a_{3}+\cdots+a_{1990}\right) \\
N=\left(a_{1}+a_{2}+a_{3}+\cdots+a_{1990}\right)\left(a_{2}+a_{3}+\cdots+a_{1989}\right)
\end{array}
$$

Then the relationship between $M$ and $N$ is
(A) $M=N$.
(B) $M>N$.
(C) $M<N$.
(D) The relationship is uncertain.
(China Beijing High School Mathematics Competition, 1990)
[Solution] Let $A=a_{2}+a_{3}+\cdots+a_{1989}$, then
$$
M=\left(a_{1}+A\right)\left(A+a_{1990}\right), \quad N=\left(a_{1}+A+a_{1990}\right) A \text {. }
$$

and $\quad M=a_{1} A+a_{1990} A+A^{2}+a_{1} a_{1990}$,
also
$$
\begin{array}{l}
N=a_{1} A+a_{1990} A+A^{2} \text {, } \\
\because a_{i}>0(i=1,2, \cdots, 1990), \quad \therefore \quad M>N \text {. } \\
\end{array}
$$

Therefore, the correct choice is $(C)$.