Let $ p,q,r$ be distinct real numbers that satisfy: $ q\equal{}p(4\minus{}p), \, r\equal{}q(4\minus{}q), \, p\equal{}r(4\minus{}r).$ Find all possible values of $ p\plus{}q\plus{}r$.
1. Let \( p, q, r \) be distinct real numbers that satisfy the given system of equations:
   \[
   q = p(4 - p), \quad r = q(4 - q), \quad p = r(4 - r).
   \]
   We introduce new variables \( a, b, c \) such that:
   \[
   a = p - 2, \quad b = q - 2, \quad c = r - 2.
   \]
   This transforms the system into:
   \[
   b = f(a), \quad c = f(b), \quad a = f(c),
   \]
   where \( f(x) = 2 - x^2 \).

2. We need to solve the equation \( a = f(f(f(a))) \). Substituting \( f(x) = 2 - x^2 \), we get:
   \[
   f(f(f(a))) = 2 - (2 - (2 - a^2)^2)^2.
   \]
   Therefore, we need to solve:
   \[
   f(f(f(a))) - a = 0.
   \]

3. To find the roots of this equation, we note that \( f(a) = a \) gives the fixed points of \( f \):
   \[
   2 - a^2 = a \implies a^2 + a - 2 = 0 \implies (a - 1)(a + 2) = 0.
   \]
   Thus, \( a = 1 \) or \( a = -2 \).

4. We now consider the polynomial \( f(f(f(a))) - a \):
   \[
   f(f(f(a))) - a = 2 - (2 - (2 - a^2)^2)^2 - a.
   \]
   This polynomial is of degree 8. The roots of this polynomial include \( a = 1 \) and \( a = -2 \). The sum of the roots of a polynomial \( P(x) = 0 \) of degree \( n \) with leading coefficient 1 is given by Vieta's formulas as the negative of the coefficient of \( x^{n-1} \) divided by the leading coefficient. Here, the sum of the roots of \( f(f(f(a))) - a = 0 \) is zero because the coefficient of \( a^7 \) is zero.

5. The polynomial \( f(f(f(a))) - a \) factors as:
   \[
   (a - 1)(a + 2)(a^3 - 3a - 1)(a^3 - a^2 - 2a + 1) = 0.
   \]
   The roots of \( a^3 - 3a - 1 = 0 \) and \( a^3 - a^2 - 2a + 1 = 0 \) are the remaining roots.

6. We need to verify that the roots of \( a^3 - 3a - 1 = 0 \) and \( a^3 - a^2 - 2a + 1 = 0 \) form triples \( \{a_1, f(a_1), f(f(a_1))\} \) and \( \{a_2, f(a_2), f(f(a_2))\} \) respectively. The sum of the roots of each cubic polynomial is zero by Vieta's formulas.

7. Therefore, the sum of the roots of \( f(f(f(a))) - a = 0 \) is zero, and the sum of the roots of the two cubic polynomials is zero. The total sum of \( a, b, c \) is zero.

8. Finally, we need to find \( p + q + r \):
   \[
   p + q + r = (a + 2) + (b + 2) + (c + 2) = a + b + c + 6.
   \]
   Since \( a + b + c = 0 \), we have:
   \[
   p + q + r = 6.
   \]

The final answer is \(\boxed{6}\).