, 2013

Solution:
Answer: $72$

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
$$\sum_{1 \leq i<j \leq 5}\left\lfloor a_{i}+a_{j}\right\rfloor = \sum_{1 \leq i<j \leq 5} (a_{i}+a_{j}) - \{a_{i}+a_{j}\} = 80 - \sum_{1 \leq i<j \leq 5} \{a_{i}+a_{j}\},$$
so it suffices to maximize
$$\sum_{1 \leq i<j \leq 5} \{a_{i}+a_{j}\} = \sum_{i=1}^{5} \{a_{i}+a_{i+2}\} + \sum_{i=1}^{5} \{a_{i}+a_{i+1}\}$$
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.