Since your array is circular, it's best to consider indices $\mod n$, where $n$ is the length of your array. So normally you would name the elements $A_0, A_1, A_2, \ldots, A_{n-1}$, but we can also talk about $A_n$ (which is identical to $A_0$), a.s.o.
A continuous subarray now is something like $A_k,A_{k+1},\ldots, A_{k+r-1}$ with $r\in \{0,\ldots, n\}$, with $r$ the length of that subarray. Note that $r=0$ corresponds to the empty subarray, $r=n$ is the complete array.
Note that each continuous subarray A has a complement $c(A)$, containing the other elements of $A$: $c(A)=A_{k+r},A_{k+r+1},\ldots,A_{k+n-1}$, which is also a continuous sbuarray. Also note that the complement of the complement is again the original subarray.
It's obvious that the sum of elements of a continuous subarray (let's call it $s(A)$ for a subset $A$) and it's complement is $T$, because it's just all the elements of A added together.
So if you have a minimum continuous subarray $X$, I claim that it's complement $c(X)$ is a maximum continuous subarray. Why is that?
We have $s(X)+s(c(X))=T$ as noted above. If there was a continuous subarray $Y$ with $s(Y) > s(c(X))$, then we can look at the complement of $Y$! We have again $s(Y)+s(c(Y))=T$, so we get
$$ s(c(Y)) = T - s(Y) < T - s(c(X)) = s(X),$$
which means $c(Y)$ has a smaller sum of elements than $X$, which is a contradiction to the choice of $X$. So there can't be such a $Y$, so we proved that $c(X)$ is a maximum continuous subarray, which proves the result you want:
$$s(X)+s(c(X))=T$$
So your intuition is right, and this proof shows why: minimum and maximum continuous subarrays (there can be more than one) can be paired as complements of each other.
