An odd integer $ n \ge 3$ is said to be nice if and only if there is at least one permutation $ a_{1}, \cdots, a_{n}$ of $ 1, \cdots, n$ such that the $ n$ sums $ a_{1} \minus{} a_{2} \plus{} a_{3} \minus{} \cdots \minus{} a_{n \minus{} 1} \plus{} a_{n}$, $ a_{2} \minus{} a_{3} \plus{} a_{3} \minus{} \cdots \minus{} a_{n} \plus{} a_{1}$, $ a_{3} \minus{} a_{4} \plus{} a_{5} \minus{} \cdots \minus{} a_{1} \plus{} a_{2}$, $ \cdots$, $ a_{n} \minus{} a_{1} \plus{} a_{2} \minus{} \cdots \minus{} a_{n \minus{} 2} \plus{} a_{n \minus{} 1}$ are all positive. Determine the set of all `nice' integers.
1. **Understanding the Problem:**
   We need to determine the set of all odd integers \( n \geq 3 \) such that there exists a permutation \( a_1, a_2, \ldots, a_n \) of \( 1, 2, \ldots, n \) where the sums \( S_i = a_i - a_{i+1} + a_{i+2} - \cdots - a_{i+n-2} + a_{i+n-1} \) are all positive. Here, the indices are taken modulo \( n \).

2. **Assumption and Simplification:**
   Suppose \( n \) is nice and let \( A = (a_1, a_2, \ldots, a_n) \) be a permutation that satisfies the condition. Any cyclic permutation of \( A \) is also a witness, so we can assume without loss of generality that \( a_1 = 1 \).

3. **Expression for Sums:**
   Consider the sums:
   \[
   S_1 = a_1 - a_2 + a_3 - a_4 + \cdots - a_{n-1} + a_n
   \]
   \[
   S_2 = a_2 - a_3 + a_4 - a_5 + \cdots - a_n + a_1
   \]
   \[
   S_3 = a_3 - a_4 + a_5 - a_6 + \cdots - a_1 + a_2
   \]
   and so on.

4. **Simplifying \( S_1 \) and \( S_2 \):**
   Since \( a_1 = 1 \), we have:
   \[
   S_1 = 1 - a_2 + a_3 - a_4 + \cdots - a_{n-1} + a_n
   \]
   \[
   S_2 = a_2 - a_3 + a_4 - a_5 + \cdots - a_n + 1
   \]

5. **Equating Positive Sums:**
   For \( S_1 \) and \( S_2 \) to be positive, we need:
   \[
   1 + (a_3 + a_5 + \cdots + a_n) - (a_2 + a_4 + \cdots + a_{n-1}) > 0
   \]
   \[
   1 + (a_2 + a_4 + \cdots + a_{n-1}) - (a_3 + a_5 + \cdots + a_n) > 0
   \]

6. **Balancing the Sums:**
   Adding these two inequalities, we get:
   \[
   2 > 0
   \]
   which is always true. However, for the sums to be balanced, we must have:
   \[
   a_2 + a_4 + \cdots + a_{n-1} = a_3 + a_5 + \cdots + a_n
   \]

7. **Sum of All Elements:**
   The sum of all elements from 1 to \( n \) is:
   \[
   1 + 2 + \cdots + n = \frac{n(n+1)}{2}
   \]
   Since \( n \) is odd, \( n+1 \) is even, making the sum an integer. For the sums to be equal, the total sum must be even, implying \( n(n+1) \) must be even. Since \( n \) is odd, \( n+1 \) is even, so \( n(n+1) \) is always even.

8. **Modulo Condition:**
   For the sums to be balanced, the total sum must be divisible by 4. Since \( n(n+1) \) is even, we need:
   \[
   \frac{n(n+1)}{2} \equiv 0 \pmod{2}
   \]
   This simplifies to:
   \[
   n(n+1) \equiv 0 \pmod{4}
   \]
   Since \( n \) is odd, \( n \equiv 1 \pmod{4} \).

9. **Constructing a Permutation:**
   To complete the proof, we need to construct a permutation for \( n \equiv 1 \pmod{4} \). For example, for \( n = 5 \):
   \[
   (1, 3, 5, 2, 4)
   \]
   This permutation satisfies the condition as:
   \[
   S_1 = 1 - 3 + 5 - 2 + 4 = 5
   \]
   \[
   S_2 = 3 - 5 + 2 - 4 + 1 = -3 + 3 = 1
   \]
   and so on.

The final answer is \( \boxed{ n \equiv 1 \pmod{4} } \).