Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?
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[*]$P(x)$ has no rational root;
[*]For every positive integer $n$, there exists an integer $m$ such that $n$ divides $P(m)$.
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1. **Step 1: Constructing the Polynomial**

   We consider the polynomial \( P(x) = (x^3 + 3)(x^8 - 16) \). We need to show that this polynomial satisfies the given conditions:
   - \( P(x) \) has no rational root.
   - For every positive integer \( n \), there exists an integer \( m \) such that \( n \) divides \( P(m) \).

2. **Step 2: Showing \( P(x) \) has no rational root**

   To show that \( P(x) \) has no rational root, we use the Rational Root Theorem. The Rational Root Theorem states that any rational root of a polynomial with integer coefficients must be a factor of the constant term divided by a factor of the leading coefficient.

   For \( P(x) = (x^3 + 3)(x^8 - 16) \):
   - The constant term of \( x^3 + 3 \) is 3.
   - The constant term of \( x^8 - 16 \) is -16.
   - The leading coefficient of both \( x^3 + 3 \) and \( x^8 - 16 \) is 1.

   Therefore, any rational root of \( P(x) \) must be a factor of \( 3 \times (-16) = -48 \). The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48 \).

   We check these values:
   - \( x^3 + 3 \) has no rational roots because \( x^3 = -3 \) has no rational solutions.
   - \( x^8 - 16 \) has no rational roots because \( x^8 = 16 \) implies \( x = \pm 2^{1/4} \), which are not rational.

   Therefore, \( P(x) \) has no rational roots.

3. **Step 3: Showing divisibility for every positive integer \( n \)**

   Let \( n = \prod_{i=1}^N p_i^{\alpha_i} \), where \( p_1, p_2, \ldots, p_N \) are distinct primes and \( \alpha_i \) are their respective exponents. By the Chinese Remainder Theorem, if we can find integers \( m_i \) such that \( p_i^{\alpha_i} \mid P(m_i) \), then we can find an integer \( m \) such that \( n \mid P(m) \).

   Hence, it suffices to show this claim for prime powers.

4. **Step 4: Handling odd prime powers**

   Consider \( x^8 - 16 = (x^2 - 2)(x^2 + 2)(x^4 + 4) = (x^2 - 2)(x^2 + 2)((x-1)^2 + 1)((x+1)^2 + 1) \).

   For any odd prime \( p \), at least one of \( -2, 2, -1 \) is a quadratic residue modulo \( p \). Therefore, there exists an \( x \) such that \( p \mid x^8 - 16 \). Since \( x \not\equiv 0 \pmod{p} \) (as \( p > 2 \)), by Hensel's Lemma, for any prime \( p > 2 \) and exponent \( k \), there exists an \( x \) such that \( p^k \mid x^8 - 16 \).

5. **Step 5: Handling powers of 2**

   Consider \( x^3 + 3 \). We claim that for any \( k \), there exists an \( x \) such that \( 2^k \mid x^3 + 3 \).

   We show this by induction on \( k \):
   - For \( k = 1 \), \( x^3 + 3 \equiv 0 \pmod{2} \) if \( x \equiv 1 \pmod{2} \).
   - Assume for \( k = j \), there exists \( x \) such that \( 2^j \mid x^3 + 3 \). Then \( x^3 + 3 = 2^j y \) for some integer \( y \).
   - Consider \( x' = x + 2^j \). Then \( (x')^3 + 3 = (x + 2^j)^3 + 3 = x^3 + 3 + 3x^2 \cdot 2^j + 3x \cdot (2^j)^2 + (2^j)^3 \).
   - Since \( 2^j \mid x^3 + 3 \), we have \( 2^j \mid 3x^2 \cdot 2^j + 3x \cdot (2^j)^2 + (2^j)^3 \).
   - Therefore, \( 2^{j+1} \mid (x')^3 + 3 \).

   By induction, for any \( k \), there exists \( x \) such that \( 2^k \mid x^3 + 3 \).

6. **Conclusion**

   Combining Steps 4 and 5 into Step 1, we conclude that for every positive integer \( n \), there exists an integer \( m \) such that \( n \mid P(m) \).

The final answer is \( \boxed{ P(x) = (x^3 + 3)(x^8 - 16) } \)