9. (16 points) Let the real-coefficient polynomial
$$
P_{i}(x)=x^{2}+b_{i} x+c_{i}\left(b_{i}, c_{i} \in \mathbf{R}, i=1,2, \cdots, n\right)
$$

be distinct, and for any $1 \leqslant i<j \leqslant n, P_{i}(x)+$ $P_{j}(x)$ has exactly one real root. Find the maximum value of $n$.
9. The maximum value of $n$ is 3.
On the one hand, when $n=3$, take
$$
\begin{array}{l}
P_{1}(x)=x^{2}-4, P_{2}(x)=x^{2}-4 x+6, \\
P_{3}(x)=x^{2}-8 x+12 .
\end{array}
$$

Then $P_{1}(x)+P_{2}(x)=2(x-1)^{2}$,
$$
\begin{array}{l}
P_{1}(x)+P_{3}(x)=2(x-2)^{2}, \\
P_{3}(x)+P_{2}(x)=2(x-3)^{2},
\end{array}
$$

satisfying the conditions of the problem.
On the other hand, when $n \geqslant 4$, assume
$$
P_{i}(x)+P_{j}(x)=2\left(x-t_{i j}\right)^{2},
$$

where $t_{i j} \in \mathbf{R}, i=1,4, j=2,3$.
Then for any real number $x$, we have
$$
\begin{array}{l}
\sum_{i=1}^{4} P_{i}(x)=2\left(x-t_{12}\right)^{2}+2\left(x-t_{43}\right)^{2} \\
=2\left(x-t_{13}\right)^{2}+2\left(x-t_{42}\right)^{2} .
\end{array}
$$

Thus, $t_{12}+t_{43}=t_{13}+t_{42}$, and
$$
t_{12}^{2}+t_{43}^{2}=t_{13}^{2}+t_{42}^{2} \text {. }
$$

Hence $\left(t_{12}, t_{43}\right)=\left(t_{13}, t_{42}\right)$ or $\left(t_{12}, t_{43}\right)=\left(t_{24}, t_{13}\right)$
$\Rightarrow P_{i}(x)(i=1,2,3,4)$ are the same, which is a contradiction.
Therefore, the maximum value of $n$ is 3.