Solution:
Suppose there are four numbers $a, b, c, d$ among the given nine numbers which leave the same remainder modulo $20$. Then $a+b \equiv c+d \pmod{20}$ and we are done.

If not, there are two possibilities:

(1) We may have two disjoint pairs $\{a, c\}$ and $\{b, d\}$ obtained from the given nine numbers such that $a \equiv c \pmod{20}$ and $b \equiv d \pmod{20}$. In this case we get $a+b \equiv c+d \pmod{20}$.

(2) Or else there are at most three numbers having the same remainder modulo $20$ and the remaining six numbers leave distinct remainders which are also different from the first remainder (i.e., the remainder of the three numbers). Thus there are at least $7$ distinct remainders modulo $20$ that can be obtained from the given set of nine numbers. These $7$ remainders give rise to $\binom{7}{2} = 21$ pairs of numbers. By pigeonhole principle, there must be two pairs $\left(r_1, r_2\right), \left(r_3, r_4\right)$ such that $r_1 + r_2 \equiv r_3 + r_4 \pmod{20}$. Going back we get four numbers $a, b, c, d$ such that $a+b \equiv c+d \pmod{20}$.

If we take the numbers $0, 0, 0, 1, 2, 4, 7, 12$, we check that the result is not true for these eight numbers.