[Solution] The sum of the squares of six odd numbers must be even, so we exclude $(D)$.
Let these six odd numbers be $2 k_{1}+1, 2 k_{2}+1, \cdots 2 k_{6}+1$. Note that
$$
\begin{aligned}
& \left(2 k_{1}+1\right)^{2}+\left(2 k_{2}+1\right)^{2}+\cdots+\left(2 k_{6}+1\right)^{2} \\
= & 4\left(k_{1}^{2}+k_{2}^{2}+\cdots+k_{6}^{2}\right)+4\left(k_{1}+k_{2}+\cdots+k_{6}\right)+6 .
\end{aligned}
$$

Therefore, this number minus 6 must be divisible by 4. Among the three numbers 280, 368, and 382, only 382 minus 6 is divisible by 4.
Thus, the answer is $(C)$.
$30 \cdot 35$ Six odd numbers, their sum is 42, their sum of squares can only be
(A) 280.
(B) 368.
(C) 382.
(D) 423.
(China Jiangxi Nanchang Junior High School Mathematics Competition, 1990)