Saudi Arabia Mathematical Competitions 2012

Let $S_i$ be the sum of the first $i$ numbers in the set $M$. We have $$\begin{array}{rl}{S}_{2k}& =(1+2)+(4+5)+\cdots +(3k-2+3k-1)\\ & =(6\cdot 1-3)+(6\cdot 2-3)+\cdots +(6k-3)\\ & =3k(k+1)-3k=3k^2.\end{array}$$ Then $S_{2k+1}=S_{2k}+3k+1=3k^2+3k+1$.

Case 1. The sum of $2n$ consecutive elements of $M$ is $$a_{2k+1}+a_{2k+2}+\cdots +a_{2l}=S_{2l}-S_{2k},$$ where $n=l-k$. We obtain $3(l^2-k^2)=300$, so therefore $$(l-k)(l+k)=100,$$ giving the systems $$\begin{gathered} \{\begin{array}{l}l-k=1\\ l+k=100\end{array};\{\begin{array}{l}l-k=2\\ l+k=50\end{array};\{\begin{array}{l}l-k=4\\ l+k=25\end{array}; \\ \{\begin{array}{l}l-k=5\\ l+k=20\end{array};\{\begin{array}{l}l-k=10\\ l+k=10\end{array} \end{gathered}$$ Only the second and the last are solvable in integers, and we obtain $l=26$, $k=24$ and $l=10$, $k=0$. Thus we get $n=2$, $n=10$.

Case 2. The sum of $2n$ consecutive elements of $M$ is $$a_{2k}+a_{2k+1}+\cdots +a_{2l-1},$$ where $n=l-k$, $k\ge 1$. We obtain $3(l^2+l-k^2-k)=300$, and so $(l-k)(l+k+1)=100$. The solutions that work are $$\{\begin{array}{l}l-k=1\\ l+k=99\end{array};\{\begin{array}{l}l-k=4\\ l+k=24\end{array};\{\begin{array}{l}l-k=5\\ l+k=19\end{array}$$ We get $l=50,k=49$, $l=14,k=10$, $l=12,k=7$, meaning that $n=1,4,5$.

The solutions are $n\in \{1,2,4,5,10\}$.