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 $c \equiv d(\bmod 20)$. 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 discinct. 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 $n=9$.