How many integers $ N$ less than 1000 can be written as the sum of $ j$ consecutive positive odd integers from exactly 5 values of $ j\ge 1$?
To solve the problem, we need to determine how many integers \( N \) less than 1000 can be written as the sum of \( j \) consecutive positive odd integers from exactly 5 values of \( j \ge 1 \).

1. **Form of the Sum of Consecutive Odd Integers**:
   The sum of \( j \) consecutive positive odd integers can be expressed as:
   \[
   (i + j)^2 - i^2 = j(2i + j)
   \]
   where \( i \) is the starting odd integer and \( j \) is the number of terms.

2. **Factorization Requirement**:
   We need \( N \) to be factorable into two factors of the same parity in five ways. This means \( N \) must have exactly 5 distinct pairs \((j, k)\) such that \( N = jk \) and both \( j \) and \( k \) are either both odd or both even.

3. **Odd \( N \)**:
   For \( N \) to be odd, it must have 10 total factors (since each pair \((j, k)\) counts twice, once as \((j, k)\) and once as \((k, j)\)). The number of factors of \( N \) is given by:
   \[
   (e_1 + 1)(e_2 + 1) = 10
   \]
   where \( N = p_1^{e_1} p_2^{e_2} \).

   - **Case 1**: \( e_1 = 0, e_2 = 9 \) is impossible since \( 3^9 > 1000 \).
   - **Case 2**: \( e_1 = 1, e_2 = 4 \):
     - \( 3^4 = 81 \), \( 5^4 = 625 \), and only \( 3^4 \) can be multiplied by another prime.
     - \( \lfloor \frac{1000}{81} \rfloor = 12 \), so \( p_2 = 3 \) and \( p_1 = 5, 7, 11 \).

   - **Case 3**: \( e_1 = 2, e_2 = 2 \):
     - \( N = (3 \times 5)^2 = 225 \), \( (3 \times 7)^2 = 441 \).

   Thus, the odd \( N \) values are:
   \[
   \boxed{N = 405, 567, 891, 225, 441}
   \]

4. **Even \( N \equiv 0 \pmod{4}\)**:
   Let \( k = \frac{N}{4} \). We need \( k \) to have five distinct choices for the first factor, \(\frac{j}{2}\), which must be smaller than the second factor.

   - **Case 1**: \( (e_1 + 1)(e_2 + 1) = 10 \):
     - \( e_1 = 0, e_2 = 9 \) is impossible since \( 2^9 = 512 > 250 \).
     - \( e_1 = 1, e_2 = 4 \):
       - \( 2^4 = 16 \), \( \lfloor \frac{250}{16} \rfloor = 15 \), so \( p_2 = 2 \) and \( p_1 = 3, 5, 7, 11, 13 \).

   - **Case 2**: \( (e_1 + 1)(e_2 + 1) = 9 \):
     - \( e_1 = 2, e_2 = 2 \):
       - \( p_1 \times p_2 < \sqrt{250} \approx 15 \), giving \( p_1 \times p_2 = 6, 10, 14, 15 \).

   Thus, the \( k \) values are:
   \[
   \boxed{k = 48, 80, 112, 176, 208, 36, 100, 196, 225}
   \]

5. **Total Values of \( N \)**:
   Combining both cases, we get:
   \[
   \boxed{14}
   \]