[Answer] 151
[Solution] First, we prove that the 7 must be the last digit. If it is not the last digit, then there must be another digit after it, let's assume this digit is $A$. Then $\overline{7 A}$ should be a multiple of 17 or 23. It is easy to know $\left\{\begin{array}{l}17 \times 4=68 \\ 17 \times 5=85\end{array}\right.$, $\left\{\begin{array}{l}23 \times 3=69 \\ 23 \times 4=92\end{array}\right.$,

So there is no multiple of $\overline{7 A}$ among these numbers. Thus, we have proven that 7 must be the last digit. Then, we start to reverse from the end, $17,517,8517,68517,468517,3468517,23468517,923468517,6923468517$,
$$
46923468517,346 \underbrace{92346}_{\text {period }} 8517, \cdots \cdots .
$$

From this, we know that for a 31-digit number, excluding the last 4 digits, there are 27 digits left. Since $27=5 \times 5+2$, the final sum of all digits is $8+5+1+7+(9+2+3+4+6) \times 5+4+6=151$.
14. A 31-digit integer, if each pair of adjacent digits of this integer is considered as a two-digit number, any such two-digit number can be divisible by 17 or 23. Additionally, this 31-digit integer contains only one 7. Then the sum of all the digits of this 31-digit number is $\qquad$