［Solution］Let the four integers be $a, b, c, d$. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{1}{3}(a+b+c)+d=29 \\
\frac{1}{3}(b+c+d)+a=23 \\
\frac{1}{3}(a+c+d)+b=21 \\
\frac{1}{3}(a+b+d)+c=17 .
\end{array}\right.
$$

Solving these equations, we get $a=12, b=9, \quad c=3, d=21$. Therefore, the answer is $(B)$.
3.54 There are four integers. If any three are chosen, their average plus the other integer equals 29, 23, 21, and 17. One of the original four integers is
(A) 19.
(B) 21.
(C) 23.
(D) 29.
(E) 17.
(6th American High School Mathematics Examination, 1955)