We call the polynomial $P (x)$ simple if the coefficient of each of its members belongs to the set $\{-1, 0, 1\}$.
Let $n$ be a positive integer, $n> 1$. Find the smallest possible number of terms with a non-zero coefficient in a simple $n$-th degree polynomial with all values at integer places are divisible by $n$.

1. We need to find the smallest possible number of terms with a non-zero coefficient in a simple $n$-th degree polynomial $P(x)$ such that all values at integer places are divisible by $n$. A simple polynomial has coefficients in $\{-1, 0, 1\}$.

2. We start by noting that a polynomial with only one non-zero term cannot satisfy the condition for all integer values, as it would not be divisible by $n$ for all integers. Therefore, the number of non-zero terms must be at least 2.

3. We will now construct a polynomial with exactly 2 non-zero terms and show that it satisfies the given condition. Consider the polynomial $P(x) = x^n - x$. 

4. Evaluate $P(x)$ at any integer $k$:
   \[
   P(k) = k^n - k
   \]
   We need to show that $P(k)$ is divisible by $n$ for all integers $k$.

5. By Fermat's Little Theorem, for any integer $k$ and any prime $p$ dividing $n$, we have:
   \[
   k^p \equiv k \pmod{p}
   \]
   This implies that $k^n \equiv k \pmod{p}$ for any prime $p$ dividing $n$.

6. Since $n$ can be factored into primes as $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, we have:
   \[
   k^n \equiv k \pmod{p_i^{e_i}}
   \]
   for each prime power $p_i^{e_i}$ dividing $n$.

7. By the Chinese Remainder Theorem, since $k^n \equiv k \pmod{p_i^{e_i}}$ for all $i$, it follows that:
   \[
   k^n \equiv k \pmod{n}
   \]
   Therefore, $k^n - k$ is divisible by $n$ for any integer $k$.

8. Thus, the polynomial $P(x) = x^n - x$ satisfies the condition that all values at integer places are divisible by $n$ and has exactly 2 non-zero terms.

The final answer is $\boxed{2}$.