For how many triples of positive integers $(x,y,z)$, there exists a positive integer $n$ such that $\dfrac{x}{n} = \dfrac{y}{n+1} = \dfrac{z}{n+2}$ where $x+y+z=90$?

$ 
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 5
\qquad\textbf{(C)}\ 6
\qquad\textbf{(D)}\ 7
\qquad\textbf{(E)}\ 9
$
To solve the problem, we need to find the number of triples \((x, y, z)\) of positive integers such that there exists a positive integer \(n\) satisfying the equation:

\[
\frac{x}{n} = \frac{y}{n+1} = \frac{z}{n+2}
\]

and the condition:

\[
x + y + z = 90
\]

1. Let \(k\) be the common value of the fractions. Then we can write:
   \[
   x = kn, \quad y = k(n+1), \quad z = k(n+2)
   \]

2. Substitute these expressions into the sum condition:
   \[
   kn + k(n+1) + k(n+2) = 90
   \]

3. Simplify the equation:
   \[
   k(n + n + 1 + n + 2) = 90
   \]
   \[
   k(3n + 3) = 90
   \]
   \[
   k \cdot 3(n + 1) = 90
   \]
   \[
   k(n + 1) = 30
   \]

4. Since \(k\) and \(n+1\) are positive integers, \(n+1\) must be a divisor of 30. The positive divisors of 30 are:
   \[
   1, 2, 3, 5, 6, 10, 15, 30
   \]

5. Each divisor \(d\) of 30 corresponds to \(n+1 = d\), so:
   \[
   n = d - 1
   \]

6. We need to check that \(n\) is positive:
   \[
   n = d - 1 > 0 \implies d > 1
   \]

7. Therefore, the valid divisors of 30 that make \(n\) positive are:
   \[
   2, 3, 5, 6, 10, 15, 30
   \]

8. This gives us 7 valid values for \(n\), corresponding to 7 triples \((x, y, z)\).

Thus, the number of such triples is:

\[
\boxed{7}
\]