$a,a+1,a+2,a+3,a+4,a+5$, and $a+6$ comprise a complete set of residues modulo $7$,
so we can take $a=0$ without loss of generality.
For $n=1$, the sum is congruent to $0+1+2+3+4+5+6\equiv0\bmod7$.
For $n=2$, the sum is congruent to $0+1+4+2+2+4+1\equiv0\bmod7$.
For $n=3$, the sum is congruent to $0+1+1-1+1-1-1\equiv0\bmod7$.
For $n=4$, the sum is congruent to $0+1+2+4+4+2+1\equiv0\bmod7$.
For $n=5$, the sum is congruent to $0+1+4+5+2+3+6\equiv0\bmod7$.
For $n=6$, the sum is congruent to $0+1+1+1+1+1+1\equiv6\bmod7$.
For $n=6k+m$, the sum is congruent to that for $m$, as a consequence of Fermat's little theorem.
Therefore, the sum is a multiple of $7$ unless $n$ is a multiple of $6$.
Notice that both $79$ and $83$ are primes.
Thus, they would each fit into $(m)(m+1)(m+2)$, and thus two or one of these would be a factor of each term in the brackets. Moreover, see how the terms have a difference of at most $2$ ($m$ and $m+2$). Thus, we can rewrite the term that contains a factor of $79$ as $79a$, and the term that contains a factor of $83$ as $83b$. Now upon subtracting them, we have the difference between the two terms to be at most two.
Thus, we have that $$|79a - 83b| \leq 2$$
Now, I will list out, in ascending order, the first multiples of $79$ and $83$, where $79$ is red and $83$ is black.
$\color{red}{79},83,\color{red}{158},166,\color{red}{237},{249},\color{red}{316},{332},\color{red}{395},415,\color{red}{474},498,\color{red}{553},581,\color{red}{632},664,\color{red}{711},747,\color{red}{790},830,\color{red}{869},913,\color{red}{948},996,\color{red}{1027},1079,\color{red}{1106},1162,\color{red}{1185},1245,\color{red}{1264},1328,\color{red}{1343},1411,\color{red}{1422},1494,\color{red}{1501},1577,\color{red}{1580},\color{red}{1659},1660$
It is easily verified that the difference of $2$ or less only appears at $1659$ and $1660$. Thus, the smallest values for $a$ and $b$ are $a=21$ and $b=20$.
Since they have a difference of one, we can put $1660$ and $1659$ as $m$ and $(m-1)$ respectively, or $(m-1)$ and $(m-2)$ respectively. However, we can see that the former option gives us $m=1660$ but the latter option gives us $m=1661$, thus we will pick the former option and yield that $m=1660$.
Best Answer
Suppose there are two of them: $2^r a$ and $2^r b$ say with $a$ and $b$ odd and $a<b$. Then $b\ge a+2$, and the list of consecutive numbers also contains $2^r(a+1)$. But $a+1$ is even.