Find all polynomials $p(x)$ with real coefficients that have the following property: there exists a polynomial $q(x)$ with real coefficients such that $$p(1) + p(2) + p(3) +\dots + p(n) = p(n)q(n)$$ for all positive integers $n$. 
1. **Consider the case when \( P(x) \) is a constant polynomial:**
   - Let \( P(x) \equiv c \) where \( c \) is a constant.
   - The given equation becomes:
     \[
     nc = cQ(n) \implies \forall n \in \mathbb{N}: c(Q(n) - n) = 0
     \]
   - This implies \( c = 0 \) or \( Q(n) = n \) for all \( n \).
   - Therefore, the solutions in this case are:
     \[
     P(x) \equiv 0 \quad \text{with} \quad Q(x) \in \mathbb{R}[x]
     \]
     or
     \[
     P(x) \equiv c \quad \text{with} \quad Q(x) \equiv x
     \]

2. **Assume \( P(x) \) is not constant:**
   - Write \( P(x) \) and \( Q(x) \) in their general forms:
     \[
     P(x) = a_s x^s + a_{s-1} x^{s-1} + \dots + a_0
     \]
     \[
     Q(x) = b_t x^t + b_{t-1} x^{t-1} + \dots + b_0
     \]
   - The given equation can be rewritten as:
     \[
     \forall n \geq 2: P(n-1)Q(n-1) = P(n)(Q(n) - 1)
     \]
   - Since this holds for infinitely many \( n \), it must hold for all real numbers \( n \).

3. **Analyze the degrees of \( P(x) \) and \( Q(x) \):**
   - If \( \deg(P) > 1 \) and \( \deg(Q) > 1 \), consider the leading terms:
     \[
     [x^{t+s-1}] = a_s b_{t-1} + a_{s-1} b_t - s a_s b_t - t a_s b_t = a_s b_{t-1} + a_{s-1} b_t
     \]
   - This leads to:
     \[
     (s + t) a_s b_t = 0
     \]
   - Since \( a_s \neq 0 \) and \( b_t \neq 0 \), this is a contradiction. Therefore, we must have \( \deg(P) \leq 1 \) or \( \deg(Q) \leq 1 \).

4. **Assume \( \deg(P) \leq 1 \):**
   - Since \( P \) is not constant, \( \deg(P) = 1 \).
   - Let \( P(x) = a_1 x + a_0 \).

5. **Analyze \( \deg(Q) \):**
   - If \( \deg(Q) > 1 \), consider the leading terms:
     \[
     [x^t] = a_1 b_{t-1} + a_0 b_t - t a_1 b_t = a_1 b_{t-1} + a_0 b_t
     \]
   - This leads to a contradiction. Therefore, \( \deg(Q) \leq 1 \).

6. **Assume \( \deg(Q) = 1 \):**
   - Let \( Q(x) = bx + c \).
   - The given equation becomes:
     \[
     P(x)(Q(x) - 1) = P(x-1)Q(x-1)
     \]
   - Substitute \( Q(x) = bx + c \):
     \[
     P(x)(bx + c - 1) = P(x-1)(b(x-1) + c)
     \]
   - Equate the coefficients of \( x^s \):
     \[
     a_s (c - b) + a_{s-1} b - s a_s b = a_s (c - 1) + a_{s-1} b
     \]
   - This simplifies to:
     \[
     a_s b (s + 1) = a_s \implies b = \frac{1}{s + 1}
     \]

7. **Form of \( P(x) \):**
   - Write \( P(x) \) as:
     \[
     P(x) = \alpha (x - r_1)(x - r_2) \dots (x - r_s)
     \]
   - Plugging into the equality \( P(x)(Q(x) - 1) = P(x-1)Q(x-1) \) and solving for \( r_1 \) and \( r_s \):
     \[
     Q(r_s + 1) = 1 \quad \text{and} \quad Q(r_1 - 1) = 0
     \]
   - This implies:
     \[
     r_s - r_1 = s - 1
     \]
   - Therefore, \( P(x) \) has the form:
     \[
     P(x) = \alpha (x - r)(x - r - 1) \dots (x - r - s + 1)
     \]
   - Check the conditions \( P(1) = 0 \) or \( Q(1) = 1 \):
     \[
     r + k = 1 \quad \text{or} \quad r = -s + 1
     \]
   - Thus, \( r \in \{1, 0, \dots, -s + 1\} \).

8. **Conclusion:**
   - The polynomials \( P(x) \) that satisfy the given condition are:
     \[
     P(x) = \alpha (x - r)(x - r - 1) \dots (x - r - s + 1)
     \]
     where \( \alpha \in \mathbb{R} \) and \( r \in \{1, 0, \dots, -s + 1\} \).
   - The corresponding polynomial \( Q(x) \) is:
     \[
     Q(x) = \frac{x + 1 - r}{s + 1}
     \]

The final answer is \( \boxed{ P(x) = \alpha (x - r)(x - r - 1) \dots (x - r - s + 1) } \) where \( \alpha \in \mathbb{R} \) and \( r \in \{1, 0, \dots, -s + 1\} \).