16. (UKR) Determine the smallest integer \( n \geq 4 \) for which one can choose four different numbers \( a, b, c \), and \( d \) from any \( n \) distinct integers such that \( a+b-c-d \) is divisible by 20.
16. Let $S$ be a set of integers such that for no four distinct elements $a, b, c, d \in S$, it holds that $20 \mid a+b-c-d$. It is easily seen that there cannot exist distinct elements $a, b, c, d$ with $a \equiv b$ and $c \equiv d \pmod{20}$. Consequently, if the elements of $S$ give $k$ different residues modulo 20, then $S$ itself has at most $k+2$ elements. Next, consider these $k$ elements of $S$ with different residues modulo 20. They give $\frac{k(k-1)}{2}$ different sums of two elements. For $k \geq 7$ there are at least 21 such sums, and two of them, say $a+b$ and $c+d$, are equal modulo 20; it is easy to see that $a, b, c, d$ are distinct. It follows that $k$ cannot exceed 6, and consequently $S$ has at most 8 elements. An example of a set $S$ with 8 elements is $\{0,20,40,1,2,4,7,12\}$. Hence the answer is $n=9$.