7. (10 points) 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 . $\qquad$
【Analysis】This problem can be solved by setting the greatest common divisor as a parameter, finding the value of the parameter, and then obtaining the value of $n$.
【Solution】Solution: Let the greatest common divisor of $4 n+5$ and $7 n+6$ be $k$, then $(4 n+5) \div k$ is an integer, $(7 n+6) \div k$ is an integer,

To eliminate $n$ by taking the difference, the left expression is multiplied by 7, and the right expression is multiplied by 4, the results are still integers, then $[7(4 n+5)-4(7 n+6)] \div k=11 \div k$ is an integer,

Since $k \neq 1$, then $11 \div k$ is an integer only when $k$ is 11, meaning the greatest common divisor of the two algebraic expressions greater than 1 is 11. Also, because $[2(4 n+5)-(7 n+6)] \div k$ is an integer, [here $(4 n+5)$ is multiplied by 2 to take the difference to make the coefficient of $n$ 1 for easier selection]

Substituting $k=11$, we have $(n+4) \div 11$ is an integer
Since $n<50$
Then $n=7,18,29,40$.
$$
7+18+29+40=94 \text {. }
$$

Therefore, the sum of all possible values of $n$ is 94.
The answer is: 94.