For any natural number $n$, connect the origin $O$ and the point $A_{n}(n, n+3)$. Let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Try to find: $f(1)+f(2)+\cdots+f(1996)$.
The equation of the line segment $O A_{n}$ is $y=\frac{n+3}{n} x(0 \leqslant x \leqslant n)$. Therefore, $f(x)$ is the number of integer solutions of this equation in the interval $(0, n)$. When $n=3 k$ (where $k$ is a positive integer), the equation becomes
$$
y=\frac{k+1}{k} x(0<x<3 k) \text {. }
$$

It has two sets of integer solutions
$$
\left\{\begin{array} { l } 
{ x = k , } \\
{ y = k + 1 }
\end{array} \text { and } \left\{\begin{array}{l}
x=2 k, \\
y=2(k+1) .
\end{array}\right.\right.
$$

When $n=3 k \pm 1$ (where $k$ is a non-negative integer), it is easy to see that $n$ and $n+3$ are coprime, so the equation has no integer solutions satisfying $0<x<n$.
From the above discussion, we get
$$
f(n)=\left\{\begin{array}{ll}
0 & \text { when } n=3 k \pm 1, \\
2 & \text { when } n=3 k.
\end{array}\right.
$$

Therefore,
$$
\begin{array}{l}
f(1)+f(2)+\cdots+f(1996) \\
=2 \times\left[\frac{1996}{3}\right]=2 \times 665=1330 .
\end{array}
$$
(Shi Li Lu, Henan Provincial Traffic School, 450052)