The answer is $n=2^k-1$ for some integer $k\geq 1$.
There is a bijection between mediocre subsets of $\{1,\ldots,n\}$ and
mediocre subsets of $\{2,\ldots,n+1\}$ given by adding $1$ to each
element of the subset; thus $A(n+1)-A(n)$ is the number of mediocre
subsets of $\{1,\ldots,n+1\}$ that contain $1$. It follows that
$A(n+2)-2A(n+1)+A_n = (A(n+2)-A(n+1))-(A(n+1)-A(n))$ is the difference
between the number of mediocre subsets of $\{1,\ldots,n+2\}$ containing
$1$ and the number of mediocre subsets of $\{1,\ldots,n+1\}$ containing
$1$. This difference is precisely the number of mediocre subsets of
$\{1,\ldots,n+2\}$ containing both $1$ and $n+2$, which we term
"mediocre subsets containing the endpoints." Since $\{1,\ldots,n+2\}$
itself is a mediocre subset of itself containing the endpoints, it
suffices to prove that this is the only mediocre subset of
$\{1,\ldots,n+2\}$ containing the endpoints if and only if $n=2^k-1$ for
some $k$.

If $n$ is not of the form $2^k-1$, then we can write $n+1 = 2^a b$ for
odd $b>1$. In this case, the set $\{1+m b \, | \, 0 \leq m \leq 2^a\}$
is a mediocre subset of $\{1,\ldots,n+2\}$ containing the endpoints: the
average of $1+m_1 b$ and $1+m_2 b$, namely $1+\frac{m_1+m_2}{2} b$, is
an integer if and only if $m_1+m_2$ is even, in which case this average
lies in the set.

It remains to show that if $n=2^k-1$, then the only mediocre subset of
$\{1,\ldots,n+2\}$ containing the endpoints is itself. This is readily
seen by induction on $k$. For $k=1$, the statement is obvious. For
general $k$, any mediocre subset $S$ of $\{1,\ldots,n+2=2^k+1\}$
containing $1$ and $2^k+1$ must also contain their average, $2^{k-1}+1$.
By the induction assumption, the only mediocre subset of
$\{1,\ldots,2^{k-1}+1\}$ containing the endpoints is itself, and so $S$
must contain all integers between $1$ and $2^{k-1}+1$. Similarly, a
mediocre subset of $\{2^{k-1}+1,\ldots,2^k+1\}$ containing the endpoints
gives a mediocre subset of $\{1,\ldots,2^{k-1}+1\}$ containing the
endpoints by subtracting $2^{k-1}$ from each element. By the induction
assumption again, it follows that $S$ must contain all integers between
$2^{k-1}+1$ and $2^k+1$. Thus $S = \{1,\ldots,2^k+1\}$ and the induction
is complete.