7. Let $n$ be a natural number less than 50, then the sum of all possible values of $n$ for which $4 n+5$ and $7 n+6$ have a common divisor greater than 1 is $\qquad$ .


Let's analyze: Suppose the greatest common divisor greater than 1 of $4 n+5$ and $7 n+6$ is $A$, then $A|(4 n+5), A|(7 n+6)$. Multiplying $(4 n+5)$ by 7 and $(7 n+6)$ by 4, and then subtracting to eliminate $n$, the difference is divisible by 11, $(4 n+5) \times 7 - (7 n+6) \times 4 = 11$, and 11 is a prime number, so $A$ can only be 11. $(4 n+5)$ and $(7 n+6)$ are both multiples of 11. To find all $n$, $2 \times (4 n+5) - (7 n+6) = n + 4$, and $11 \mid (n + 4)$, so $n = 7, 18, 29, 40$. Therefore, the answer is $7 + 18 + 29 + 40 = 94$.