[Solution] If the curve passes through the integer point $(a, b)$, then $b=\frac{1}{5}\left(a^{2}-a+1\right)$.
That is, $a^{2}-a+1-5 b=0$, viewing it as a quadratic trinomial in $a$, it has integer roots, so $\Delta=1-4(1-5 b)=20 b-3=0$ or is a perfect square.
$\because b$ is an integer, $20 b-3 \neq 0$;
Also, since $b$ is an integer, $20 b-3$ ends in 7, thus it is not a perfect square (perfect squares end only in $0,1,4,5,6,9$).
$\therefore$ The curve $y=\frac{1}{5}\left(x^{2}-x+1\right)$ does not pass through any integer points. Hence, the answer is $(A)$.
37 -10 Points in the plane with both coordinates as integers are called integer points. Then the number of integer points through which the curve $y=$ $\frac{1}{5}\left(x^{2}-x+1\right)$ passes is
(A) 0.
(B) 1.
(C) 2.
(D) Infinitely many.
(High School Mathematics Competition in Huanggang Region, Hubei Province, China, 1989)