, 2012

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

Case 1. The sum of $2n$ consecutive elements of $M$ is
$$a_{2k+1} + a_{2k+2} + \dots + a_{2l} = S_{2l} - S_{2k},$$
where $n = l - k$. We obtain $3(l^2 - k^2) = 300$, so therefore
$$(l-k)(l+k) = 100,$$
giving the systems
$$\begin{cases} l - k = 1 \\ l + k = 100 \end{cases} \quad ; \quad \begin{cases} l - k = 2 \\ l + k = 50 \end{cases} \quad ; \quad \begin{cases} l - k = 4 \\ l + k = 25 \end{cases} ; \\ \begin{cases} l - k = 5 \\ l + k = 20 \end{cases} ; \quad \begin{cases} l - k = 10 \\ l + k = 10 \end{cases}$$
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
$$a_{2k} + a_{2k+1} + \dots + a_{2l-1},$$
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
$$\begin{cases} l - k = 1 \\ l + k = 99 \end{cases} \quad ; \quad \begin{cases} l - k = 4 \\ l + k = 24 \end{cases} \quad ; \quad \begin{cases} l - k = 5 \\ l + k = 19 \end{cases}$$
We get
$l = 50, k = 49$, $l = 14, k = 10$, $l = 12, k = 7$,
meaning that $n = 1, 4, 5$.

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