Given that Keith is keeping an equal number of square and rectangular boxes, we denote the number of boxes of each type as \( n \).
For the square boxes, they can be stacked in groups of 7, which implies that \( n \) must be a multiple of 7. Therefore, we write: \[ n = 7k \quad \text{for some integer } k \]
For the rectangular boxes, they can be stacked in groups of 3, which implies that \( n \) must also be a multiple of 3. We can express this as: \[ n = 3m \quad \text{for some integer } m \]
Since \( n \) needs to satisfy both conditions, we need to find the smallest positive integer \( n \) that is a multiple of both 7 and 3. To accomplish this, we look for the least common multiple (LCM) of 7 and 3.
The method to find the LCM involves identifying the prime factors:
- The prime factorization of 7 is simply \( 7^1 \).
- The prime factorization of 3 is \( 3^1 \).
The LCM is computed by taking the highest power of each prime factor: \[ \text{LCM}(7, 3) = 7^1 \cdot 3^1 = 21 \]
Thus, \( n = 21 \) is the smallest number that is both a multiple of 7 and a multiple of 3. This means that Keith could have a minimum of 21 square boxes and 21 rectangular boxes.
Therefore, the smallest number of each type of box that Keith could have is:
\[ \boxed{21} \]