16. Find the smallest integer \( n (n \geq 4) \), such that from any \( n \) integers, four different numbers \( a, b, c, d \) can be selected, making \( a + b - c - d \) divisible by 20.
Solution: First, we consider the different residue classes modulo 20. For a set with $k$ elements, there are $\frac{1}{2} k(k - 1)$ integer pairs. If $\frac{1}{2} k(k - 1) > 20$, i.e., $k \geq 7$, then there exist two pairs $(a, b)$ and $(c, d)$ such that $a + b \equiv c + d \pmod{20}$, and $a, b, c, d$ are all distinct.

In general, consider a set of 9 different elements. If this set has 7 or more elements belonging to different residue classes modulo 20, then from the previous derivation, we can find four distinct numbers $a, b, c, d$ such that $a + b \equiv c + d \pmod{20}$. Assuming that in this set, at most six elements belong to the same residue class modulo 20, we can still find $a, b, c, d$ such that $a + b - c - d$ is divisible by 20.

For a set of 8 elements, such as
$$
\{0, 20, 40, 1, 2, 4, 7, 12\},
$$

we prove that it does not satisfy the required property. The residues of these numbers modulo 20 are $0, 0, 0, 1, 2, 4, 7, 12$. These residues have the property that each number is greater than the sum of any two smaller numbers, and the sum of any two numbers is less than 20. Let $a, b, c, d$ be four distinct numbers from this set, treating 20 and 40 as 0. Assume $a$ is the largest of the four chosen numbers. Then $a + b - c - d < a + b - c < a + b < 20$. Therefore, $a + b - c - d$ is not divisible by 20.

The minimum value of $n$ is 9.