4. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y \leqslant 3 x, \\
y \geqslant \frac{1}{3} x, \\
x+y \leqslant 100
\end{array}\right.
$$

is
$$
1+2+3+\cdots+101=\frac{102 \times 101}{2}=5151 \uparrow .
$$

The region enclosed by $y=\frac{1}{3} x, x+y=100$, and the $x$-axis (excluding the boundary $y=\frac{1}{3} x$) contains
$$
\begin{aligned}
& (1+1+1+1)+(2+2+2+2)+\cdots \\
& +(25+25+25+25) \\
= & 4 \times(1+2+\cdots+25)=1300 \text { points. }
\end{aligned}
$$

The region enclosed by $y=3 x, x+y=100$, and the $y$-axis (excluding the boundary $y=3 x$) also contains 1300 points. Therefore, the total number of points satisfying the given conditions is $5151 - 2 \times 1300 = 2551$.
4. 2551 .
The region enclosed by the two coordinate axes and the line $x+y=100$ (including the boundaries) contains