, 2020

Solution:
We make use of the identity
$$\sum_{i=0}^{\ell} F_{i}=F_{\ell+2}-1$$
(easily proven by induction) which implies
$$\sum_{i=k}^{\ell} F_{i}=F_{\ell+2}-F_{k+1}$$
Applying this several times yields
$$\begin{aligned} & \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100} \sum_{i_{5}=0}^{100} F_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}} \\ = & \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100}\left(F_{i_{1}+i_{2}+i_{3}+i_{4}+102}-F_{i_{1}+i_{2}+i_{3}+i_{4}+1}\right) \\ = & \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100}\left(F_{i_{1}+i_{2}+i_{3}+204}-2 F_{i_{1}+i_{2}+i_{3}+103}+F_{i_{1}+i_{2}+i_{3}+2}\right) \\ = & \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100}\left(F_{i_{1}+i_{2}+306}-3 F_{i_{1}+i_{2}+205}+3 F_{i_{1}+i_{2}+104}-F_{i_{1}+i_{2}+3}\right) \\ = & \sum_{i_{1}=0}^{100}\left(F_{i_{1}+408}-4 F_{i_{1}+307}+6 F_{i_{1}+206}-4 F_{i_{1}+105}+F_{i_{1}+4}\right) \\ = & F_{510}-5 F_{409}+10 F_{308}-10 F_{207}+5 F_{106}-F_{5} . \end{aligned}$$
This representation is unique because the Fibonacci terms grow exponentially quickly, so e.g. the $F_{510}$ term dominates, forcing $n_{1}=510$ and similarly for the other terms. The final answer is
$$510+409+308+207+106+5=1545$$