Understanding Factorial Zeroes
The number of trailing zeroes in \( N! \) is determined by the formula:
\[
\text{Zeroes} = \left\lfloor \frac{N}{5} \right\rfloor + \left\lfloor \frac{N}{25} \right\rfloor + \left\lfloor \frac{N}{125} \right\rfloor + \ldots
\]
This counts the factors of 5 in \( N! \), as there are always more factors of 2 than 5.

Given Condition
- Let \( z(N) \) be the number of trailing zeroes in \( N! \).
- The problem states:
- \( z(N) = x \)
- \( z(N + 3) = x + 2 \)
From the condition, we derive that adding 3 to \( N \) introduces exactly 2 additional factors of 5.

Condition Analysis
- For \( z(N + 3) \) to be \( z(N) + 2 \):
- \( N \) must be such that when we increment it by 3, we hit a new multiple of 5 or multiple of 25.
This can occur in two scenarios:
1. \( N + 3 \equiv 0 \mod 5 \) but not \( N \equiv 0 \mod 5 \)
2. \( N + 3 \equiv 0 \mod 25 \)

Finding Valid \( N \)
- As \( N \) is a two-digit number (from 10 to 99), we check each \( N \):
- **For \( N \equiv 2 \mod 5 \)**:
- \( N = 12, 17, 22, \ldots, 97 \) (total 18 values)
- **For \( N \equiv 22 \mod 25 \)**:
- Valid values are \( N = 22, 47, 72, 97 \) (total 4 values)

Total Valid \( N \)
- The two conditions overlap for \( N \equiv 2 \mod 5 \).
- Therefore, the total count of valid \( N \) values is **18**.
In conclusion, the total number of two-digit integers \( N \) satisfying the given condition is **18**.