Let $(s_n) = \{s_1,s_2,\dots,s_n\}$ be a sequence of real numbers. Let us assume that we do not know the the value of any $s_i, \ i\in \{1,2,\dots,n\}$, but instead the sequence $(s_n)$ has been divided in some unknown way to $k$ non-overlapping subsequences and let the length of $j$th, $j\in \{1,2,\dots,k\}$ such a subsequence be $l_j$. We can assume that all lengths $l_j$ are greater than zero and thus smaller than $n$.
Assume we have been given only the set of means of these subsequences $(a_k) = \{a_1,a_2,\dots,a_k\}$, e.g. $a_1 = \frac{s_2+s_5+s_{n-2}}{3},$ thus $l_1 = 3.$
Question: is it possible to use only the values of $(a_k)$ and $n$ to find out the mean of $(s_n)$?
If $k = n$ or if $k=1$, the question is trivial. If $l_j$ are equal for all values of $j$, the mean of $(a_k)$ is equal to the mean of $(s_n)$.
Let us find all possible ways to represent $n$ as a sum of $k$ integers, call such a representation a set of weights. For example, if $n=5$ and $k=3$, $(1,2,2)$ is a set of weights. If we consider every possible way to multiply the values of $(a_k)$ with corresponding values in a set of weights, add them together and divide the result by $n$, at least one of these is the true mean of $(s_n)$, namely the one with the set of weights $(l_1,l_2,\dots,l_k)$. However, we can then only say that the true mean is in this list of possibilities found using this method.
No, without the weights it is impossible. If you have the weights, then the answer is:
$$\frac{1}{n}\sum_{i=1}^ka_il_i$$
But if you don't have the weights, then the averages no longer contain enough information. This is alluded to in the final paragraph. For example, consider the sequences $1, 1, 2, 2$ and $1, 1, 1, 2$. We could split the first into $1, 1$ and $2, 2$ and the second into $1, 1, 1$ and $2$. In both cases, we get $k=2$ and $(a_1, a_2)$ = $(1, 2)$. But clearly the averages of the two sequences differ, so $k$, $n$ and the vector of $a_i$ is clearly not enough information to determine the average of the initial sequence. Entropy has won.