$4 \cdot 218$ In a middle school mathematics competition, three problems $A$, $B$, and $C$ were given. Among the 25 students who participated in the competition, each student solved at least one problem. Among those who did not solve $A$, the number of students who solved $B$ is twice the number of students who solved $C$. The number of students who solved only $A$ is one more than the number of students who solved $A$ among the remaining students. Among the students who solved only one problem, half did not solve $A$. How many students solved only $B$?
[Solution]. Let the number of people who solved not only $A$ be $x$, only $B$ be $y$, and did not solve $A$ but solved $B$ and $c$ be $z$. Then (refer to the diagram on the previous page),
$$
\left\{\begin{array}{l}
x+x+1+y+z+\frac{y-z}{2}=25, \\
x+1=y+\frac{y-z}{2} .
\end{array}\right.
$$

Solving these equations, we get
$$
\left\{\begin{array}{l}
y=\frac{25-x}{3}, \\
z=23-3 x .
\end{array}\right.
$$

From $y \geqslant z$, we know that $x \geqslant 6$,
and only when $x=7$, $y$ and $z$ are both positive integers,
so $x=7, z=2, y=6$.
Therefore, the number of people who only solved $B$ is 6.