It is known that there are $n$ integers $a_1, a_2, \cdots, a_n$ such that
$$a_1 + a_2 + \cdots + a_n = 0 \qquad \text{and} \qquad a_1 \times a_2 \times \cdots \times a_n = n.$$
Determine all possible values of $n$.
1. **Establishing the necessity of \(4 \mid n\):**

   Given the conditions:
   \[
   a_1 + a_2 + \cdots + a_n = 0 \quad \text{and} \quad a_1 \times a_2 \times \cdots \times a_n = n
   \]
   we need to determine the possible values of \(n\).

   - **Step 1: \(a_i \neq 0\):**
     Since \(a_1 \times a_2 \times \cdots \times a_n = n \neq 0\), none of the \(a_i\) can be zero.

   - **Step 2: \(n\) is odd:**
     Suppose \(n\) is odd. Then all \(a_i\) must be odd because the product of an even number of integers is even, and \(n\) is odd. However, the sum of an odd number of odd integers is odd, which contradicts the given sum \(a_1 + a_2 + \cdots + a_n = 0\) (an even number). Therefore, \(n\) cannot be odd.

   - **Step 3: \(n = 2t\) where \(t\) is odd:**
     Suppose \(n = 2t\) where \(t\) is odd. Since the product \(a_1 \times a_2 \times \cdots \times a_n = n\) is even but not a multiple of 4, there must be exactly one even \(a_i\) and the rest odd. However, the sum of an odd number of odd integers plus one even integer is odd, which again contradicts the given sum \(a_1 + a_2 + \cdots + a_n = 0\). Therefore, \(n\) cannot be of the form \(2t\) where \(t\) is odd.

   - **Conclusion:**
     From the above steps, \(n\) must be a multiple of 4. 

2. **Showing that \(n = 4k\) works:**

   Suppose \(n = 4k\). We need to construct \(a_1, a_2, \ldots, a_n\) such that their sum is 0 and their product is \(n\).

   - **Case 1: \(n = 2^a b\) where \(a \geq 2\) and \(b\) is odd:**
     - If \(a\) is even, take \(a_1 = 2, a_2 = -2, \ldots, a_{a-1} = 2, a_a = -2, a_{a+1} = b, a_{a+2} = -1, a_{a+3} = -1, \ldots, a_{a+b-1} = -1\). The remaining \(2^a b - a - b + 1\) terms are even, so we can alternate between \(+1\) and \(-1\) to ensure the sum is 0.
     - If \(a\) is odd, take \(a_1 = 4, a_2 = -2, \ldots, a_{a-1} = -2, a_a = 2, a_{a+1} = b, a_{a+k} = -1\) for \(k = 2, \ldots, a + b + 1\). Again, the remaining terms can be alternated between \(+1\) and \(-1\) to ensure the sum is 0.

   - **Verification:**
     In both cases, the sum of the \(a_i\) is 0, and the product is \(n\), satisfying the given conditions.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) must be a multiple of 4.