Let $P(3\cos\alpha, 2\sin\alpha)$, $\alpha \neq 0$, be a point on the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, $Q$ be a point on the circle $x^2+y^2-14x-14y+82=0$ and $R$ be a point on the line $x+y=5$ such that the centroid of the triangle $PQR$ is $\left(2+\cos\alpha, 3+\dfrac{2}{3}\sin\alpha\right)$. Then the sum of the ordinates of all possible points $R$ is:

Let the coordinates of the points be $P(3\cos\alpha, 2\sin\alpha)$, $Q(x_Q, y_Q)$, and $R(x_R, y_R)$.

The centroid of $\triangle PQR$ is given by:
$\left( \dfrac{3\cos\alpha + x_Q + x_R}{3}, \dfrac{2\sin\alpha + y_Q + y_R}{3} \right)$

We are given that the centroid is $\left(2+\cos\alpha, 3+\dfrac{2}{3}\sin\alpha\right)$. Equating the coordinates, we get:
$\dfrac{3\cos\alpha + x_Q + x_R}{3} = 2 + \cos\alpha \implies x_Q + x_R = 6$
$\dfrac{2\sin\alpha + y_Q + y_R}{3} = 3 + \dfrac{2}{3}\sin\alpha \implies y_Q + y_R = 9$

Since the point $R$ lies on the line $x+y=5$, we have $x_R = 5 - y_R$.

Substituting $x_R$ into the equation for $x_Q$:
$x_Q = 6 - x_R = 6 - (5 - y_R) = y_R + 1$
And we already have $y_Q = 9 - y_R$.

The point $Q(x_Q, y_Q)$ lies on the circle $x^2+y^2-14x-14y+82=0$. Rewriting the circle's equation in standard form:
$(x-7)^2 + (y-7)^2 = 16$

Substitute $x_Q$ and $y_Q$ into the circle's equation:
$(y_R + 1 - 7)^2 + (9 - y_R - 7)^2 = 16$
$(y_R - 6)^2 + (2 - y_R)^2 = 16$

Expanding the terms:
y_R^2 - 12y_R + 36 + y_R^2 - 4y_R + 4 = 16
2y_R^2 - 16y_R + 40 = 16
2y_R^2 - 16y_R + 24 = 0
y_R^2 - 8y_R + 12 = 0

Factoring the quadratic equation:
$(y_R - 2)(y_R - 6) = 0$

Thus, the possible ordinates for point $R$ are $y_R = 2$ and $y_R = 6$.

The sum of the ordinates of all possible points $R$ is $2 + 6 = 8$.

Answer: $8$