We claim that the minimum is 72. This can be achieved by taking $a_{1}=a_{2}=a_{3}=a_{4}=0.4$ and $a_{5}=18.4$. To prove that this is optimal, note that $\sum_{1 \leq i<j \leq 5}\left\lfloor a_{i}+a_{j}\right\rfloor=\sum_{1 \leq i<j \leq 5}\left(a_{i}+a_{j}\right)-\left\{a_{i}+a_{j}\right\}=80-\sum_{1 \leq i<j \leq 5}\left\{a_{i}+a_{j}\right\}$ so it suffices to maximize $\sum_{1 \leq i<j \leq 5}\left\{a_{i}+a_{j}\right\}=\sum_{i=1}^{5}\left\{a_{i}+a_{i+2}\right\}+\sum_{i=1}^{5}\left\{a_{i}+a_{i+1}\right\}$ where $a_{6}=a_{1}$ and $a_{7}=a_{2}$. Taking each sum modulo 1, it is clear that both are integers. Thus, the above sum is at most $2 \cdot 4=8$, and our original expression is at least $80-8=72$, completing the proof.