Answer: 12

Solution:

Since $13! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13,$ a valid $m = 2^x 3^y 5^z 7^w 11^u 13^v$ must leave every exponent of $\frac{13!}{m}$ even: $x \in \{0, 2, 4, 6, 8, 10\},$ $y \in \{1, 3, 5\},$ $z \in \{0, 2\},$ and $w = u = v = 1.$

The choices are independent, so the sum of all such $m$ factors as $$\begin{aligned} &(1 + 4 + \cdots + 1024) \\ &\quad {}\cdot (3 + 27 + 243) \\ &\quad {}\cdot (1 + 25) \cdot 7 \cdot 11 \cdot 13 \\ &= 1365 \cdot 273 \cdot 26 \cdot 1001. \end{aligned}$$ Since $1365 = 3 \cdot 5 \cdot 7 \cdot 13,$ $273 = 3 \cdot 7 \cdot 13,$ $26 = 2 \cdot 13,$ and $1001 = 7 \cdot 11 \cdot 13,$ the sum equals $2^1 3^2 5^1 7^3 11^1 13^4,$ and $a + b + c + d + e + f$ $= 1 + 2 + 1 + 3 + 1 + 4$ $= 12.$