We will show that for each $i=3,4,...,2016$, there exist sequence $a_1,a_2,...,a_i$ of positive integers such that $\sum_{t=r}^{t=s}{a_t}$ is composite number for all $1\leq r\leq s\leq i$ and $gcd(a_j,a_{j+1})=1$ for all $j=1,2,...,i-1$ and $gcd(a_j,a_{j+2})=1$ for all $j=1,2,...,i-2$ For $i=3$, we choose sequence $4,81,121$ Suppose there exist sequence $a_1,a_2,...,a_i$, then we choose $a_{i+1}=p^c$ where $p$ is prime number larger than $\Big( \sum_{t=1}^{t=i}{a_t}\Big)+1$ and $c$ is a positive integer such that $\prod_{t=1}^{t=i}{(q_t-1)}\mid c$ where $q_t$ is a prime divisor of $\Big( \sum_{l=t}^{l=i}{a_l}\Big) +1$ for all $t=1,2,...,i$ This value of $a_{i+1}$ will give us for all $t=1,2,...,i$, $\Big( \sum_{l=t}^{l=i}{a_l}\Big) +p^c\equiv_{q_t} \Big( \sum_{l=t}^{l=i}{a_l}\Big)+1\equiv_{q_t} 0 $ and $a_{i+1}>q_t$ And we clearly have $gcd(a_t,a_{i+1})=gcd(a_t,p^c)=1$ for all $t=1,2,...,i$

The answer is yes, and prove that, it suffices to give an example satisfying the conditions in part (b); so we'll do just that. Pick an odd prime $p$, and another bunch of odd primes (pairwise distinct and also distinct from $p$ ), say, $p_1,p_2,\cdots ,p_{2016},q_1,q_2,\cdots ,q_{2016}$. Now by virtue of CRT, choose a number $a_0$ such that: \begin{align*} a_0 &\equiv 1\pmod{2}\\ a_0 &\equiv 1\pmod{p}\\ &\text{and...}\\ a_0+2ip &\equiv 0\pmod{p_iq_i} \text{ for }i=1,2,3,\cdots ,2016.\end{align*}Now set $a_i=a_0+2ip$ for $i=1,2,\cdots ,2016$. We claim that this works. Indeed, since the $a_i$'s are in an arithmetic progression, we have $$a_{r} + a_{r+1} + . . . + a_{s-1} + a_{s}=\left(\frac{a_r+a_s}{2}\right)\cdot(s-r+1).$$Now $\frac{a_r+a_s}{2}$ is always an integer, because by construction, $a_0$ and hence all $a_i$'s are odd. For $r\ne s$, the above sum is obviously composite, and for $r=s$, the above sum reduces to $a_r$, which is composite because $p_rq_r|a_r$. Now it remains to check that GCD condition. Fortunately, this is easy: $\gcd\left(a_i,a_{i+2}\right)=\gcd\left(a_i,a_i+4p\right)=\gcd\left(a_i,4p\right)$, which can't be $>1$ since we've been careful to set $a_i\equiv a_0\equiv 1\pmod{2}\text{ and }\pmod{ p}$. The case $\gcd\left(a_i,a_{i+1}\right)$ is similar, so we win. $\blacksquare$

My solution at the contest a) and b) $3^3;5^3$ and so on

@muuratjann I think that your solution is incorrect because if you take $r=s$ then you have that sum is $a_{r}=1^3$ which is obviously incorrect since $1$ is not a composite number.

http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Seniors_ENG_Solutions.pdf

Somebany wrote: @muuratjann I think that your solution is incorrect because if you take $r=s$ then you have that sum is $a_{r}=1^3$ which is obviously incorrect since $1$ is not a composite number. Starting from $3^3$

Does this work for the part b): Take $a_1=p_1^{2017}$ , then $a_2 \equiv 0 \pmod {p_2} $ , $a_2 \equiv 1 \pmod {a_1} $ and $a_2 \equiv -{a_1} \pmod {p_{1,2}} $ . Then inductively construct the $a_i$'s for $i \geq 3$ as follows. Take $a_s \equiv 0 \pmod {p_s} $ , $a_s \equiv -{ a_1+a_2+\cdots+a_{s-1}} \pmod {p_{k,s}} $ for $1 \leq k \leq {s-1}$ and $a_s \equiv 1 \pmod {a_{s-1} a_{s-2}} $ . This can be done by CRT as we can take sufficiently large $p_i$'s and $p_{k,i}$'s to ensure that they are pairwise coprime and also coprime to $a_{s-1}a_{s-2}$ Continue till $s=2016$.

Claim:The sum of any $r\ge 1$ consecutive terms of a arithmetic progression whose each term is odd composite number is a composite number. proof.If we take even number of consecutive terms then the sum is even.Now consider the sum of odd number of consecutive terms.Let the terms be $x,x+d,X+2d\dots x+(n-1)d$ where n is odd.Clearly the sum $ n[x+\frac{n-1}{2}d]$ is composite.$\blacksquare$ Claim:For all integer $n$ there is a arithmetic progression with $n$ terms such that the following holds: Each term is a odd composite number. GCD of any term of this AP is 1. proof.Take $M\ge (2020n)!$.Consider the AP, $M!+n!+1,M!+n!+1+(n!),M!+n!+1+2\times n!,\dots M!+n!+1+(n-1)n!$. In short,the AP with first term $M!+n!+1$ and common difference $n!$ and first $n$ terms. Obviously each term is odd composite.If a prime $p$ devides GCD of 2 terms, say $p|GCD(M!+n!+1+r(n!),M!+n!+1+s\times n!)\\ \implies p|(r-s)n!\\ \implies p|n!\\ \implies p|M!+n!+r(n!)\\ \implies p|1$. A contradiction. The third line follows from the fact that $p$ devides at least one of $|r-s|$ or $n!$ but in both cases,as $|r-s|<n$,$p|n!$.The 4th line follows from the fact that $n!|M!+n!+r(n!)$. Hence GCD of any 2 term is 1.$\blacksquare$ Taking $n=2016$ by the above 2 claims both (a),(b) holds .$\blacksquare$

The answer is positive for both cases, so proving second one is enough. Let $a_i=N!+2i+1$ for all $1\le i\le 2016$, where $N \geq 3+5+\cdots 4033$. $\gcd(a_i,a_{i+1})=\gcd(a_i,a_{i+1}-a_i)=\gcd(a_i,2)=1$ and similarly $\gcd(a_i,a_{i+2})=\gcd(a_i,4)=1$. And since $(2r+1)+(2r+3)+\cdots (2s+1)\mid N!$ for all $1 \le r \le s \le 2016$, we get $(2r+1)+(2r+3)+\cdots (2s+1)\mid a_{r} + a_{r+1} + . . . + a_{s-1} + a_s$ and since $2r+1\geq 3$ we are done!

The answer is yes for both (a) and (b) We prove this for all $n$. (Goal being $n=2016$) For $n=1$ just take $a_1=4$ or anything similar. Now for $n \rightarrow n+1$ we choose $n$ primes $p_{n,1}, p_{n,2}, \dots p_{n,n}$ such that they are all distinct and coprime to any subarray sum we have so far. Now we construct $a_{n+1}$ using $C.R.T$ such that: \[ a_{n+1} \equiv -\sum_{i=j}^n a_i \mod p_{n,j}^2\]for all $1 \leq j \leq n$ and \[ a_{n+1} \equiv 1 \mod a_na_{n-1} \]This is valid, since each of the squares are coprime to all elements, including $a_na_{n-1}$. Since each new subarray containing $a_{n+1}$ is divisible by some square, they will not be prime. Additionally, $a_{n+1}$ is coprime to both $a_n$ and $a_{n-1}$, so the construction is valid for $n+1$ and by induction for all $n$, including 2016