To find the maximum possible value of their greatest common divisor $\gcd (a_1,a_2,\cdots, a_7)$. The sum of seven distinct positive integers $a_1,a_2,\cdots, a_7$ is $315$. To find the maximum possible value of their greatest common divisor $\gcd (a_1,a_2,\cdots, a_7)$.

If they were all equal then $a_i = 45$ and the maximum possible value of their greatest common divisor would have been $45$.
Since they are distinct $\gcd(a_1,a_2,\cdots, a_7) < 45$.
Let $ \gcd(a_1,a_2,\cdots, a_7) = d$. Hence $a_i = db_i$ for some $b_i$ for all $i$.
Since $a_1 + \cdots + a_7 = 315$, we have $d(b_1 +\cdots+ b_7) = 315$ and hence $d$ divides $315$.
Thus $d$ can be any factor of $315$ smaller than $45$.

How to proceed after that?
 A: If $d$ divides both $a$ and $b$, then $d$ divides their sum.  By similar reasoning, the max gcd must be a divisor of $315 = 5 \times 7 \times 3^2.$
Further, set $d$ as the gcd.  Then, you must have that
$a_1, a_2, \cdots, a_7$ can be represented by
$(d \times x_1), (d \times x_2), \cdots, (d \times x_7).$
This implies that $x_1, \cdots, x_7$ must be distinct positive integers, such that
$\displaystyle x_1 + x_2 + \cdots + x_7 = \frac{315}{d}$.
Further, just as you are very limited in your choices for $d$, a factor of $315$, you are similarly limited in your choices for $~\displaystyle \frac{315}{d}.$
Trying to maximize $d$ is equivalent to trying to minimize $~\displaystyle \frac{315}{d}.$
This implies that you are trying to minimize $x_1 + x_2 + \cdots + x_7.$
However, since $x_1, \cdots, x_7$ must all be positive integers, their sum must be $\geq 1 + 2 + \cdots + 7 = 28.$
Then, you want $~\displaystyle \frac{315}{d}$ to be as small as possible, and still be $\geq 28.$
With $315$ being factored by $5 \times 7 \times 3^2,$ it is immediate that $\displaystyle \frac{315}{d}$ can not be an element in $\{28,29,\cdots, 34\}.$
Therefore, $~\displaystyle \frac{315}{d}$ is minimized by setting it to $(35)$.  This implies that $d = 9$, so $(9)$ is the largest possible gcd.
It remains to verify that the equation $x_1 + x_2 + \cdots + x_7 = 35$ 
does have a solution in distinct positive integers.
Since there is the obvious solution of $(x_1, x_2, \cdots, x_6, x_7) = (1,2,\cdots, 6, 14)$, the problem is solved.
