To solve this problem, we need to find the sum of possible integral values of n for which the 6th term in the expansion of (3/2 x/3)^n is numerically greatest when x = 3.

Let's first understand the expansion of the given expression.

Expansion of (3/2 x/3)^n using the binomial theorem is given by:

(3/2 x/3)^n = C(n,0) (3/2)^n (x/3)^0 + C(n,1) (3/2)^(n-1) (x/3)^1 + C(n,2) (3/2)^(n-2) (x/3)^2 + ...

The kth term in the expansion is given by:

T(k) = C(n,k) (3/2)^(n-k) (x/3)^k

To find the 6th term, we set k = 5:

T(6) = C(n,5) (3/2)^(n-5) (x/3)^5

Now, we need to find the value of n for which the 6th term is numerically greatest when x = 3.

To determine this, we need to compare the coefficients of the 6th term for different values of n.

Let's consider two consecutive values of n: n and n+1.

For n, the coefficient of the 6th term is C(n,5) (3/2)^(n-5) (x/3)^5.

For n+1, the coefficient of the 6th term is C(n+1,5) (3/2)^(n+1-5) (x/3)^5.

To find the condition for the 6th term to be numerically greatest, we equate the two coefficients and solve for n:

C(n,5) (3/2)^(n-5) (x/3)^5 = C(n+1,5) (3/2)^(n+1-5) (x/3)^5

C(n,5) = C(n+1,5)

Using the formula for binomial coefficient C(n,k) = n! / (k! * (n-k)!), we can rewrite the equation as:

n! / (5! * (n-5)!) = (n+1)! / (5! * (n+1-5)!)

Simplifying the equation, we get:

n! / (5! * (n-5)!) = (n+1) * n * (n-1) * (n-2) * (n-3) / 120

Cancelling out the common terms and simplifying further, we have:

1 / (n-4) = (n+1)(n)(n-1)(n-2)(n-3) / 120

120 = (n+1)(n)(n-1)(n-2)(n-3)(n-4)

Now, we need to find the sum of possible integral values of n that satisfy this equation.