If the quadratic equation $(\lambda+2)x^2-3\lambda x+4\lambda=0$, $\lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:

For the quadratic equation $(\lambda+2)x^2-3\lambda x+4\lambda=0$ to have two positive roots, the following three conditions must be satisfied:

1. Discriminant $D \ge 0$
$(-3\lambda)^2 - 4(\lambda+2)(4\lambda) \ge 0$
$9\lambda^2 - 16\lambda^2 - 32\lambda \ge 0$
$-7\lambda^2 - 32\lambda \ge 0$
$7\lambda^2 + 32\lambda \le 0$
$\lambda(7\lambda + 32) \le 0$
$\lambda \in \left[-\dfrac{32}{7}, 0\right]$

2. Sum of roots $> 0$
$\alpha + \beta = \dfrac{3\lambda}{\lambda+2} > 0$
$\lambda \in (-\infty, -2) \cup (0, \infty)$

3. Product of roots $> 0$
$\alpha \beta = \dfrac{4\lambda}{\lambda+2} > 0$
$\lambda \in (-\infty, -2) \cup (0, \infty)$

Taking the intersection of all three conditions, we get:
$\lambda \in \left[-\dfrac{32}{7}, -2\right)$

Since $-\dfrac{32}{7} \approx -4.57$, the possible integral values of $\lambda$ in this interval are $-4$ and $-3$.

Thus, there are $2$ possible integral values of $\lambda$.

Answer: $2$