We have given 81 real numbers $a_1, a_2, \ldots ,a_{81}$ satisfying $(1) \;\; \prod_{i=1}^{81} a_i \neq 0$, $(2) \;\; \sum_{i=1}^{81} a_i = S < 0$, $(3) \;\; a_i + a_{i+1} > 0, \; 1 \leq i \leq 80$. Assume $1 \leq k \leq 41$. Then according to (3) we have $0 < \sum_{i=1}^{k-1} (a_{2i-1} + a_{2i}) \:+\: \sum_{i=k}^{40} (a_{2i} + a_{2i+1}) = -a_{2k-1} \:+\: \sum_{i=1}^{81} a_i = S - a_{2k-1}$, i.e. $a_{2k - 1} < S$, which combined with (2) implies $(4) \;\; a_{2k - 1} < 0, \; 1 \leq k \leq 41$. Next assume $1 \leq k \leq 40$. Then by (3) and (4) $a_{2k} > -a_{2k-1} > 0$, i.e. $(5) \;\; a_{2k} > 0, \; 1 \leq k \leq 40$. Combining (4) and (5) we find that $a_1 \cdot a_2 \cdots a_{81}$ is a product of 41 negative numbers and 40 positive numbers, implying $a_1 \cdot a_2 \cdots a_{81} < 0$.