Find all polynomials $P(x)$ with integer coefficients such that for all positive integers $n$, we have that $P(n)$ is not zero and $\frac{P(\overline{nn})}{P(n)}$ is an integer, where $\overline{nn}$ is the integer obtained upon concatenating $n$ with itself.
1. **Claim and Verification**:
   We claim that \( P(x) = x^m \) for some \( m \in \mathbb{N}_0 \). To verify, we need to check if \( P(x) = x^m \) satisfies the given conditions. 

   For \( P(x) = x^m \):
   \[
   P(n) = n^m \quad \text{and} \quad P(\overline{nn}) = (10^{s(n)}n + n)^m = (10^{s(n)} + 1)^m n^m
   \]
   where \( s(n) \) is the number of digits in \( n \). 

   Then,
   \[
   \frac{P(\overline{nn})}{P(n)} = \frac{(10^{s(n)} + 1)^m n^m}{n^m} = (10^{s(n)} + 1)^m
   \]
   Since \( 10^{s(n)} + 1 \) is an integer, \((10^{s(n)} + 1)^m\) is also an integer. Thus, \( P(x) = x^m \) satisfies the condition.

2. **Assume \( P \) satisfies the conditions**:
   Suppose \( P \) is a polynomial with integer coefficients that satisfies the given conditions. For each \( q \) that divides \( P(n) \) for some \( n \), define the set:
   \[
   S_q := \{ m \in \mathbb{Z} : q \mid P(m) \}
   \]
   If \( m \in S_q \), then \( m - q \in S_q \) because \( P(m) \equiv 0 \pmod{q} \) implies \( P(m - q) \equiv P(m) \equiv 0 \pmod{q} \).

3. **Properties of \( S_q \)**:
   For \( q \) such that \( S_q \neq \emptyset \) and \( S_q \cap \mathbb{Z} \neq \mathbb{Z} \), and \( q > 10^{\deg P} \), we claim that \( S_q \) consists only of multiples of \( q \).

4. **Concatenation and divisibility**:
   If \( n \in S_q \cap \mathbb{N} \), then:
   \[
   \overline{nn} = (10^{s(n)} + 1)n \in S_q \cap \mathbb{N}
   \]
   where \( s(n) \) is the number of digits of \( n \). 

   Since \( n \in S_q \), it implies \( n + kq \in S_q \) for all \( k \). Therefore:
   \[
   (10^{s(n+kq)} + 1)(n + kq) \in S_q
   \]
   This implies:
   \[
   (10^{s(n+kq)} + 1)n \in S_q
   \]
   Since \( \{s(n+kq) : k \in \mathbb{N}\} \) contains all sufficiently large integers, we have from Fermat's Little Theorem (FLT) that:
   \[
   n(10^t + 1) \in S_q \quad \forall t \in \mathbb{N}
   \]

5. **Contradiction and conclusion**:
   If \( n \neq 0 \), then \( 10^t \) attains at least \( \log(q+1) > \log q \ge \deg P \) values modulo \( q \). Hence, \( P \) has \( \deg P + 1 \) roots modulo \( q \) (namely \( n(10^t + 1) \) for \( t = 0, 1, \ldots, \deg P \)). Therefore, \( P \) must be identically zero modulo \( q \), which is a contradiction. Thus, \( n \in S_q \implies n = 0 \), as desired.

6. **Non-zero polynomial**:
   Since \( P \) cannot be identically zero, there exists \( K \) such that for all \( q > K \), \( S_q \) consists only of multiples of \( q \) or is empty.

7. **Factorization**:
   Write \( P(x) = x^a Q(x) \) with \( Q(0) \neq 0 \). Then \( Q \) satisfies the same condition and \( Q(0) \neq 0 \). However, note that \( 0 \in S_q \) if \( S_q \neq \emptyset \). It follows that the number of primes \( q > K \) dividing some value of \( P \) is bounded, so \( P \) has only finitely many prime divisors, which is a contradiction.

8. **Conclusion**:
   Therefore, the only polynomials \( P(x) \) that satisfy the given conditions are of the form \( P(x) = x^m \).

The final answer is \( \boxed{ P(x) = x^m } \) for some \( m \in \mathbb{N}_0 \).