We call a $5$-tuple of integers [i]arrangeable[/i] if its elements can be labeled $a, b, c, d, e$ in some order so that $a-b+c-d+e=29$. Determine all $2017$-tuples of integers $n_1, n_2, . . . , n_{2017}$ such that if we place them in a circle in clockwise order, then any $5$-tuple of numbers in consecutive positions on the circle is arrangeable.

[i]Warut Suksompong, Thailand[/i]
1. **Define k-arrangeable and k-satisfiable:**
   - A $5$-tuple of integers is called *k-arrangeable* if its elements can be labeled $a, b, c, d, e$ in some order such that $a - b + c - d + e = k$.
   - A $2017$-tuple of integers $n_1, n_2, \ldots, n_{2017}$ is called *k-satisfiable* if, when placed in a circle in clockwise order, any $5$-tuple of numbers in consecutive positions on the circle is $k$-arrangeable.

2. **Lemma 1:**
   - If a $2017$-tuple of integers is $k$-satisfiable, then all of its members are $k \pmod{2}$.

   **Proof:**
   - Let $a, b, c, d, e$ be five consecutive members of the $2017$-tuple. Then:
     \[
     a - b + c - d + e = k
     \]
     Adding all terms:
     \[
     a + b + c + d + e = k + 2(b + d)
     \]
     So:
     \[
     a + b + c + d + e \equiv k \pmod{2}
     \]
     This is symmetrical, so we may assume that $a, b, c, d, e$ is in order. Let $f$ come next in the $2017$-tuple. Then:
     \[
     b + c + d + e + f \equiv k \pmod{2}
     \]
     So:
     \[
     a \equiv f \pmod{2}
     \]
     Thus, the $2017$-tuple is cyclically periodic modulo $2$ with period $5$. But since $\gcd(2017, 5) = 1$, all of the members of the $2017$-tuple are congruent modulo $2$. Then:
     \[
     a \equiv a - b + c - d + e \equiv k \pmod{2}
     \]
     So all of the members of the $2017$-tuple are $k \pmod{2}$.

3. **Lemma 2:**
   - The number of $k$-satisfiable $2017$-tuples is the same as the number of $\left\lfloor \frac{k}{2} \right\rfloor$-satisfiable $2017$-tuples.

   **Proof:**
   - Let $\left(n_1, n_2, \ldots, n_{2017}\right)$ be a $\left\lfloor \frac{k}{2} \right\rfloor$-satisfiable $2017$-tuple. Then $\left(2n_1 + 2\left\{\frac{k}{2}\right\}, 2n_2 + 2\left\{\frac{k}{2}\right\}, \ldots, 2n_{2017} + 2\left\{\frac{k}{2}\right\}\right)$ is a $k$-satisfiable $2017$-tuple. Furthermore, this mapping is unique since $2x + 2\left\{\frac{k}{2}\right\}$ is injective.
   - Now, let $\left(n_1, n_2, \ldots, n_{2017}\right)$ be a $k$-satisfiable $2017$-tuple. By Lemma 1, $n_i - 2\left\{\frac{k}{2}\right\}$ is even, so $\frac{n_i}{2} - \left\{\frac{k}{2}\right\}$ is an integer. But then $\left(\frac{n_1}{2} - \left\{\frac{k}{2}\right\}, \frac{n_2}{2} - \left\{\frac{k}{2}\right\}, \ldots, \frac{n_{2017}}{2} - \left\{\frac{k}{2}\right\}\right)$ is a $\left\lfloor \frac{k}{2} \right\rfloor$-satisfiable $2017$-tuple. Furthermore, this mapping is unique since $\frac{x}{2} - \left\{\frac{k}{2}\right\}$ is injective.
   - Thus, there is a bijection between the $k$-satisfiable $2017$-tuples and the $\left\lfloor \frac{k}{2} \right\rfloor$-satisfiable ones, so the lemma is proven.

4. **Application of Lemma 2:**
   - For a positive integer $n$, note that the sequence defined as $a_1 = n$, $a_{i+1} = \left\lfloor \frac{a_i}{2} \right\rfloor$ for $i \geq 1$ is a sequence of non-negative integers which is strictly decreasing when it is positive. This cannot go on infinitely, so at some point, $a_i = 0$. Applying this to $n = 29$, Lemma 2 (repeatedly applied) gives that the number of $29$-satisfiable $2017$-tuples is the same as the number of $0$-satisfiable $2017$-tuples.

5. **Assume there is a $0$-satisfiable $2017$-tuple that is not identically $0$:**
   - Choose the one with the smallest possible $\left|n_1\right| + \left|n_2\right| + \ldots + \left|n_{2017}\right|$. By Lemma 1, $\left(\frac{n_1}{2}, \frac{n_2}{2}, \ldots, \frac{n_{2017}}{2}\right)$ is a $0$-satisfiable $2017$-tuple. But:
     \[
     \left|\frac{n_1}{2}\right| + \left|\frac{n_2}{2}\right| + \ldots + \left|\frac{n_{2017}}{2}\right| < \left|n_1\right| + \left|n_2\right| + \ldots + \left|n_{2017}\right|
     \]
     This is a contradiction, so there is none that is not identically $0$.

6. **Conclusion:**
   - Thus, there is $1$ $0$-satisfiable $2017$-tuple and thus $1$ $29$-satisfiable $2017$-tuple. Note that the trivial $\boxed{\left(\underbrace{29, 29, \ldots, 29}_{2017}\right)}$ works, so it is the only solution.