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Answer: yes. For example consider a permutation $1,100,2,99,3,98,\dots$ For odd $i$ we have $a_i+a_{i+1}=101$. It is not difficult to check that if $u$ and $v$ have the same parity (and therefore the number of summands is odd) then for all choices of $u$ and $v=u+2\ell$ all the sums $$\sum _{i=2k-1}^{2k-1+2\ell }{a}_i=101\ell +a_{2k-1+2\ell },\sum _{i=2k}^{2k+2\ell }{a}_i=101\ell +a_{2k}$$ are different! Therefore the total number of different values is at least $51\cdot 50=2550>2500$.