1. Assigning the minimum required problems:
- We need at least 6 combinatorics problems.
- We need at least 1 problem each from algebra, geometry, and number theory.
- This accounts for $6 + 1 + 1 + 1 = 9$ problems.

2. Remaining problems:
- Since the total number of problems is 15, the remaining number of problems to be assigned is $15 - 9 = 6$.

3. Using stars and bars:
- We need to distribute these 6 remaining problems among the four subjects: algebra, geometry, combinatorics, and number theory.
- Let $a$, $g$, $c$, and $n$ represent the number of additional problems assigned to algebra, geometry, combinatorics, and number theory, respectively.
- We need to solve the equation $a + g + c + n = 6$ where $a, g, c, n \geq 0$.

4. Applying the stars and bars theorem:
- The number of non-negative integer solutions to the equation $a + g + c + n = 6$ is given by the binomial coefficient:
$$\binom{6 + 4 - 1}{4 - 1} = \binom{9}{3}$$

5. Calculating the binomial coefficient:
$$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$

Therefore, the number of non-similar somewhat evenly distributed mock AIMEs that Noew can write is $84$.