$37 \cdot 3$ On the coordinate plane, points with both coordinates as integers are called integer points. We use $I$ to denote the set of all lines, $M$ to denote the set of lines passing through exactly one integer point, $N$ to denote the set of lines not passing through any integer points, and $P$ to denote the set of lines passing through infinitely many integer points. Then, the expressions
(1) $M \cup N \cup P=I$,
(2) $N \neq \varnothing$,
(3) $M \neq \varnothing$,
(4) $P \neq \varnothing$,

the number of correct expressions is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(China High School Mathematics League, 1988)
[Solution] Let $k$ be an irrational number, then the line $y=k x+\frac{1}{2}$ does not pass through any integer points $(\because$ when $x$ is an integer, $y$ is irrational or $\frac{1}{2}$ ), hence $N \neq \varnothing$, i.e., (2) is correct.
The line $y=k x$ has only one integer point $(0,0)$, hence $M \neq \varnothing$.
The line $x=0$ clearly has infinitely many integer points, hence $P \neq \varnothing$, thus, (2), (3), and (4) are all correct.
Next, we prove that (1) holds. For any line $l$, if $l$ does not pass through any integer points, then $l \in N$;
if $l$ passes through exactly one integer point, then $l \in M$;
if $l$ passes through 2 or more than 2 integer points, then $l \in P$.
Because if $(a, b),(c, d)$ are two integer points on $l$, if $a=c$, then the equation of $l$ is $x=a$, at this time $l \in P$.

If $a \neq c$, then the equation of $l$ is $y=\frac{d-b}{c-a}(x-a)+b$, when $x=n(c-a)$ $+a$ ( $n$ is an integer), $y$ is also an integer, hence $l \in P$. This proves that (1) is correct.
In summary: All 4 expressions are correct.
Therefore, the answer is $(D)$.