Let $2^k$ be the maximal power of $2$ dividing one of the chosen numbers; since the numbers are consecutive, $2^k$ divides exactly one of them. Then one of the two l.c.m.'s under consideration is divisible by $2^k$, while the other is not.

Suppose the contrary. Consider the powers of $2$ dividing the chosen numbers; let $2^k$ be the largest among them. If at least two of the chosen numbers are divisible by $2^k$, then two consecutive such numbers will differ by $2^k$. Therefore, one of them will be divisible by $2^{k+1}$, which is impossible by the choice of $k$. Thus, among the chosen numbers, exactly one is divisible by $2^k$.

The l.c.m. of the group containing this number will be divisible by $2^k$, and the l.c.m. of the remaining group will not be. Therefore, the sum of these l.c.m.'s is not divisible by $2^k$; on the other hand, this sum is greater than $2^k$. Therefore, this sum cannot be a power of $2$.