$14 \cdot 34$ Calculate the value of the sum $\sum_{n=0}^{502}\left[\frac{305 n}{503}\right]$.
(1st China Northeast Three Provinces Mathematics Invitational Competition, 1986)
[Solution] Since 503 is a prime number,
for $n=0,1,2, \cdots, 502$, $\frac{305 n}{503}$ is not an integer.
Since $\frac{305 n}{503}+\frac{305(503-n)}{503}=305$,
the sum of the fractional parts of $\frac{305 n}{503}$ and $\frac{305(503-n)}{503}$ is 1, thus
$$
\begin{array}{l}
{\left[\frac{305 n}{503}\right]+\left[\frac{305(503-n)}{503}\right]=304 \text {. }} \\
\sum_{n=0}^{502}\left[\frac{305 n}{503}\right]=\sum_{n=1}^{502}\left[\frac{305 n}{503}\right] \\
=\sum_{n=1}^{251}\left(\left[\frac{305 n}{503}\right]+\left[\frac{305\left(\frac{503}{503}-n\right)}{5}\right]\right) \\
=304 \cdot 251 \\
=76304 . \\
\end{array}
$$