Consecutive integers in $S_n:=\{\sum_{i=1}^{n}x_i^n\ \mid\ \left(x_1,...,x_n\right)\in\mathbb{N}^n\}$ Define for $n≥2$ the set $S_n$ to be $S_n:=\{\sum_{i=1}^{n}x_i^n\ \mid\ \left(x_1,...,x_n\right)\in\mathbb{N}^n\}$ where $\mathbb{N}=\{1,2,3,...\}$. What is the longest sequence of consecutive integers in $S_n$?
For $S_2$ I wrote a short C++ program to check the set $\{x_1^2+x_2^2\ \mid\ 1≤x_1,x_2≤2000\}$ and the longest sequence I could find had a length of $3$, starting with $72,73,74$. I have a feeling that there are no for consecutive integers in the set, but I fail to see how to prove it.
For $S_3$, the same program found a sequence of $6$ consecutive integers, namely $925035,925036,925037,925038,925039,925040$, where it searched for $1≤x_1,x_2,x_3≤100$. It seems that as the $6$-tuplet is pretty big in comparison with the range searched, we keep on stumbling upon such sequences, whereas for $S_2$, it seems to stop at length $3$, because the range there was even larger. Thus conjecture that here we might find arbitrarily long such sequences.
Is this true? How to generalize to $S_n$? Are there any known results concerning these question?
Side note:
What do you think is more natural; to allow $x_i=0$ or to disallow it? From the point of view of Fermat's sum of two squares theorem (my inspiration for this question) it seems more natural to disallow it. But if we allow it, we would already find $16,17,18$ as a triplet which would be kind of nice.
Edit:
As Michael Tong remarked, a simple application of modular arithmetic allow to prove that $l_2≤3$ and $l_3≤7$. It gives a procedure to calculate an upper bound $l_n$ for any given $n$, but it isn't easily applicable in the general case.
 A: Decided to post an incomplete answer rather than keep commenting.
Since $x^2 \equiv 0,1 \pmod 4$, it is impossible for an integer $k \equiv 3 \pmod 4$ to be expressed as the sum of two squares. Hence, the bound on length for $S_2$ is $3$; this is achieved by, say, $72, 73, 74 = 6^2 + 6^2, 8^2 + 3^2, 7^2 + 5^2$ (found by the OP). Notice that the requirement $0 \in \mathbb{N}$ here is immaterial. (Cf. A082982)
Since $x^3 \equiv -1, 0, 1 \pmod 9$, it is impossible for $k \equiv 5, 6$ to be expressed as the sum of three squares. Hence the bound on length for $S_3$ is $7$. I have found $7$ consecutively: the sequence starting from $47420214$ and ending in $47420220$ can all be expressed as the sum of three non-zero positive cubes. So again, the condition $0 \in \mathbb{N}$ doesn't matter.
Now, $x^4 \equiv 0, 1 \pmod {16}$, so the longest possible sequence is $5$. This is trivially satisfied by taking $x_i = 0,1$; non-trivially we have the sequence starting with $1501248$ or $4100624$.
A: Look at the modulus: $n^n$? for $S_2$, we know that $x^2 \equiv 0,1 \mod4$, so you can't get $3 \mod 4$ as the sum of two squares.
Let $m=x^3$. Then $m \equiv 0,-1,1 \mod 9$.
Hence, let $m_1,m_2,m_3$ each be a perfect cube. Then you can't get any $m \equiv 5 \mod 9$ as a sum of $3$ cubes. 
Perhaps you could extend this argument.
