Omar made a list of all the arithmetic progressions of positive integer numbers such that the difference is equal to $2$ and the sum of its terms is $200$. How many progressions does Omar's list have?
1. **Define the arithmetic progression and its properties:**
   Let the first term of the arithmetic progression (AP) be \( a \) and the common difference be \( d = 2 \). Suppose the AP has \( k \) terms. The terms of the AP are:
   \[
   a, a+2, a+4, \ldots, a+2(k-1)
   \]

2. **Sum of the terms of the AP:**
   The sum of the first \( k \) terms of an arithmetic progression can be calculated using the formula:
   \[
   S_k = \frac{k}{2} \left(2a + (k-1) \cdot d\right)
   \]
   Substituting \( d = 2 \):
   \[
   S_k = \frac{k}{2} \left(2a + 2(k-1)\right) = \frac{k}{2} \cdot 2(a + k - 1) = k(a + k - 1)
   \]
   Given that the sum of the terms is 200, we have:
   \[
   k(a + k - 1) = 200
   \]

3. **Factorization of 200:**
   To find the possible values of \( k \) and \( a \), we need to factorize 200. The prime factorization of 200 is:
   \[
   200 = 2^3 \cdot 5^2
   \]
   The positive divisors of 200 are: \( 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200 \).

4. **Determine valid pairs \((k, a)\):**
   Since \( k \) must be a positive integer and \( a \) must also be a positive integer, we need to check each divisor \( k \) to see if \( a \) is a positive integer:
   \[
   a + k - 1 = \frac{200}{k}
   \]
   \[
   a = \frac{200}{k} - k + 1
   \]

   We need \( a \geq 1 \), so:
   \[
   \frac{200}{k} - k + 1 \geq 1
   \]
   \[
   \frac{200}{k} \geq k
   \]
   \[
   200 \geq k^2
   \]
   Therefore, \( k \leq \sqrt{200} \approx 14.14 \). The possible values of \( k \) are \( 1, 2, 4, 5, 8, 10 \).

5. **Calculate corresponding \( a \) for each valid \( k \):**
   - For \( k = 1 \):
     \[
     a = \frac{200}{1} - 1 + 1 = 200
     \]
   - For \( k = 2 \):
     \[
     a = \frac{200}{2} - 2 + 1 = 99
     \]
   - For \( k = 4 \):
     \[
     a = \frac{200}{4} - 4 + 1 = 47
     \]
   - For \( k = 5 \):
     \[
     a = \frac{200}{5} - 5 + 1 = 36
     \]
   - For \( k = 8 \):
     \[
     a = \frac{200}{8} - 8 + 1 = 18
     \]
   - For \( k = 10 \):
     \[
     a = \frac{200}{10} - 10 + 1 = 11
     \]

6. **List the arithmetic progressions:**
   The 6 progressions are:
   \[
   \begin{align*}
   &200 \\
   &99, 101 \\
   &47, 49, 51, 53 \\
   &36, 38, 40, 42, 44 \\
   &18, 20, 22, 24, 26, 28, 30, 32 \\
   &11, 13, 15, 17, 19, 21, 23, 25, 27, 29
   \end{align*}
   \]

The final answer is \(\boxed{6}\)