Let $a_1,a_2,\ldots,a_{100}$ be real numbers,each less than one,satisfy $a_1+a_2+\ldots+a_{100} >1$. Let $a_1,a_2,\ldots,a_{100}$ be real numbers,each less than one,satisfy $$a_1+a_2+\ldots+a_{100}>1$$ Prove the following statements:
(i) Let $n_0$ be the smallest integer $n$ such that  $$a_1+a_2+\ldots+a_n>1$$
Show that all the sums $a_{n_0},a_{n_0}+a_{n_0-1},\ldots,a_{n_0}+\ldots+a_1$ are positive.
(ii) Show that there exist two integers $p$ and $q,p<q$, such that the numbers $$a_q,a_q+a_{q-1},\ldots,a_q+\ldots+a_p$$  $$a_p,a_p+a_{p+1},\ldots,a_p+\ldots+a_q$$
are all positive.  
My work:
I could solve the first part by the Extremal Principle, but cannot approach the second part. I do not know extremal principal much and it is absolutely not necessary to answer using extremal principle. Please help.
 A: For part (i), let $S_k:=a_1+a_2+\cdots+a_{k}$. Then $S_{n_0-1}\le 1,\ S_{n_0}>1$. Since $S_{n_0}=S_{n_0-1}+a_{n_0}$, had $a_{n_0}$ been $\le 0$, $S_{n_0}$ would be $\le S_{n_0-1}\le 1$. Hence, $a_{n_0}>0$. Similarly, applying this argument to $S_{n_0}=S_{n_0-2}+a_{n_0-1}+a_{n_0}$ gives $a_{n_0-1}+a_{n_0}>0$ and so on. 
For part (ii), define $T_{p,q}:=\sum_{k=p}^q a_k$ and $T_{q,q}:=a_q$. Now, we know that the numbers $T_{k,n_0}\ \forall k=1,2,\cdots,n_0$ are positive. Let, $q=n_0$ and $p=\arg\max_{1\le k\le n_0}T_{k,n_0}$. Then, $T_{k,p}$ are all positive for $p\le k\le q$. Also, for all $p\le k\le q-1$,  $\ T_{p,k}=T_{p,q}-T_{k,q}\ge 0$ by definition of $p,q$. Hence the assertion is proved.
A: Let $q$ be an integer such that the sum of the first $q$ elements is maximized. It is greater than one, since if it is one then $a_q>1$. Now let $p$ be an integer such that the sum of $a_q+a_{q-1}+...a_p$ is maximized. It must be less than $q$, since if is equal that would imply that $a_q>1$. Conclude.
