25. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Given $a=\frac{p}{q}$ (where $p$ and $q$ are coprime positive integers). If the sum of all real numbers $x$ that satisfy $[x]\{x\}=a x^{2}$ is 420, then the value of $p+q$ is ( ).
(A) 245
(B) 593
(C) 929
(D) 1331
(E) 1332
25. C.

Notice that when $x<0$, $[x]\{x\}<0$. Contradiction.

Thus, we only need to consider the case where $x>0$, and $x=0$ does not affect the result.
Assume $[x]=n$. Then $\{x\}=x-n$.
Hence $[x]\{x\}=n(x-n)=a x^{2}$
$\Rightarrow a x^{2}-n x+n^{2}=0$
$\Rightarrow x=\frac{n \pm \sqrt{n^{2}-4 a n^{2}}}{2 a}$.
Consider the smallest solution of $x$, $x=\frac{n-\sqrt{n^{2}-4 a n^{2}}}{2 a}$.
Given that $[x]=n$ is the maximum value of $n$, the desired value is
$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$.
When $n=28$, $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}=406$.
Thus, $\sum_{k=1}^{28} \frac{k}{2 a}(1-\sqrt{1-4 a})$
$=\frac{1-\sqrt{1-4 a}}{2} \times 406=420$
$\Rightarrow \frac{1-\sqrt{1-4 a}}{2 a}=\frac{420}{406}=\frac{30}{29}$
$\Rightarrow a=\frac{116}{3600}=\frac{29}{900}$
$\Rightarrow p+q=929$.