If the sum of the slope and the $y$-intercept of a line is $3$, then through which point is the line guaranteed to pass?
1. Let's start by considering the equation of a line in slope-intercept form:
   \[
   y = mx + b
   \]
   where \( m \) is the slope and \( b \) is the y-intercept.

2. According to the problem, the sum of the slope \( m \) and the y-intercept \( b \) is 3:
   \[
   m + b = 3
   \]

3. We can express \( b \) in terms of \( m \):
   \[
   b = 3 - m
   \]

4. Substitute \( b = 3 - m \) into the equation of the line:
   \[
   y = mx + (3 - m)
   \]
   Simplifying, we get:
   \[
   y = mx + 3 - m
   \]

5. To find a specific point through which the line passes, we can choose a value for \( x \). Let's choose \( x = 1 \):
   \[
   y = m(1) + 3 - m
   \]
   Simplifying, we get:
   \[
   y = m + 3 - m = 3
   \]

6. Therefore, when \( x = 1 \), \( y = 3 \). This means the point \((1, 3)\) lies on the line regardless of the value of \( m \).

Conclusion:
The line is guaranteed to pass through the point \((1, 3)\).

The final answer is \( \boxed{ (1, 3) } \).