Given a real number $A$ and an integer $n$ with $2 \leq n \leq 19$, find all polynomials $P(x)$ with real coefficients such that $P(P(P(x))) = Ax^n +19x+99$.
1. Let \( P(x) \) be a polynomial with real coefficients and let \( m \) be the degree of \( P(x) \). We are given that \( P(P(P(x))) = Ax^n + 19x + 99 \).

2. The degree of \( P(P(P(x))) \) is \( m^3 \). This is because the degree of a composition of polynomials is the product of their degrees.

3. We need to consider two cases: \( A = 0 \) and \( A \neq 0 \).

### Case 1: \( A = 0 \)
4. If \( A = 0 \), then \( P(P(P(x))) = 19x + 99 \). This implies that the degree of \( P(P(P(x))) \) is 1.

5. Since \( m^3 = 1 \), we have \( m = 1 \). Therefore, \( P(x) \) must be a linear polynomial of the form \( P(x) = ax + b \).

6. Substituting \( P(x) = ax + b \) into \( P(P(P(x))) \), we get:
   \[
   P(P(P(x))) = a(a(ax + b) + b) + b = a(a^2x + ab + b) + b = a^3x + a^2b + ab + b
   \]

7. Comparing this with \( 19x + 99 \), we get:
   \[
   a^3 = 19 \quad \text{and} \quad a^2b + ab + b = 99
   \]

8. Solving \( a^3 = 19 \), we get \( a = \sqrt[3]{19} \).

9. Substituting \( a = \sqrt[3]{19} \) into \( a^2b + ab + b = 99 \), we get:
   \[
   (\sqrt[3]{19})^2b + \sqrt[3]{19}b + b = 99
   \]
   \[
   19^{2/3}b + 19^{1/3}b + b = 99
   \]
   \[
   b(19^{2/3} + 19^{1/3} + 1) = 99
   \]
   \[
   b = \frac{99}{19^{2/3} + 19^{1/3} + 1}
   \]

10. Therefore, the polynomial \( P(x) \) when \( A = 0 \) is:
    \[
    P(x) = \sqrt[3]{19}x + \frac{99}{19^{2/3} + 19^{1/3} + 1}
    \]

### Case 2: \( A \neq 0 \)
11. If \( A \neq 0 \), then the degree of \( P(P(P(x))) \) is \( n \). This implies \( m^3 = n \).

12. Given \( 2 \leq n \leq 19 \), the only integer \( m \) that satisfies \( m^3 = n \) is \( m = 2 \) and \( n = 8 \).

13. Therefore, \( P(x) \) must be a quadratic polynomial of the form \( P(x) = ax^2 + bx + c \).

14. The highest degree term in \( P(P(x)) \) is \( a^3x^4 \) and the highest degree term in \( P(P(P(x))) \) is \( a^7x^8 \).

15. Comparing this with \( Ax^8 + 19x + 99 \), we get \( a^7 = A \) and the coefficient of \( x^7 \) must be zero.

16. The coefficient of \( x^7 \) in \( P(P(P(x))) \) is \( 4a^6b \). For this to be zero, \( b \) must be zero.

17. Therefore, \( P(x) \) must be of the form \( P(x) = ax^2 + c \).

18. However, if \( P(x) \) is even, then \( P(P(P(x))) \) must also be even, which contradicts the presence of the linear term \( 19x \).

19. Hence, there is no solution if \( A \neq 0 \).

### Conclusion
20. The unique solution is when \( A = 0 \) and \( P(x) = \sqrt[3]{19}x + \frac{99}{19^{2/3} + 19^{1/3} + 1} \).

The final answer is \( \boxed{ P(x) = \sqrt[3]{19}x + \frac{99}{19^{2/3} + 19^{1/3} + 1} } \).