4. 183 Find the number of real pairs $(a, b)$ that make the system of equations $\left\{\begin{array}{l}a x+b y=1, \\ x^{2}+y^{2}=50 .\end{array}\right.$ have at least one solution, and all solutions are integer solutions.
[Solution] The graph of the equation $x^{2}+y^{2}=50$ is a circle with center at $(0,0)$ and radius $5 \sqrt{2}$, and the graph of $a x+b y=1$ is a straight line. The problem is equivalent to finding the number of lines that intersect the circle and have each intersection point as an integer point.

There are only 12 integer points on the circle: $(1,7),(1,-7),(-1,7),(-1,-7),(5$, $5),(5,-5),(-5,5),(-5,-5),(7,1),(7,-1),(-7,1),(-7,-1)$.

There are 12 lines that pass through one of these integer points and are tangent to the circle;
There are $C_{12}^{2}=66$ lines that pass through two of these integer points.
Among the above two types of lines, there are 6 lines that pass through the origin. Since lines passing through the origin cannot be written in the form $a x+b y=1$, they are not included in the original system of equations.

In summary, the number of lines represented by $a x+b y=1$ in the original system of equations can be
$$12+66-6=72 \text { (lines) }$$

Therefore, the number of real number pairs $(a, b)$ that meet the conditions in the problem is 72.