Solution:

a) A possible example for $n \geq 3$ is $e_{1} = e_{2} = 1$, $e_{i} = 0$ for $3 \leq i \leq n$. This forces $k_{2} = k_{n} = 1$ and $k_{i} = 0$ otherwise. Clearly all conditions are satisfied.

b) We prove that always $n = m + 2$. The proof proceeds in two steps.

i) We have $n \geq m + 2$: Since the number $0$ occurs in $(k_{1}, k_{2}, \ldots, k_{n})$, two elements of $(e_{1}, e_{2}, \ldots, e_{n})$ must be equal. Since, moreover, each of the $m + 1$ natural numbers from $0$ to $m$ must occur at least once, it follows that $n \geq m + 2$.

ii) We give an example for $n = m + 2 \geq 3$: Let $(e_{1}, e_{2}, \ldots, e_{n}) = (0, m, 1, m-1, \ldots)$. In this $n$-tuple the numbers are continued alternately, decreasing down to the difference $0$, so that $e_{n-1} = e_{n}$ holds. Thus every natural number from $0$ to $m$ occurs exactly once, and the number $e_{n}$ occurs exactly twice. The absolute values of the differences clearly occur once each, going downward from $m$ to $0$; the difference $|e_{n} - 0| = e_{n}$ occurs twice, because $e_{n} < m$. This establishes properties (1) and (2).