Let A be $m\times m$ matrix with all elements equal to 1 such that  $A^n={16}^{17}A,m,n\in N,$ then the sum of possible values of n.

if m=2 
 $A^1=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right],then{A}^2=\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right],A^3=\left[\begin{array}{cc}4& 4\\ 4& 4\end{array}\right],\mathrm{........}{A}^n=\left[\begin{array}{cc}{2}^{n-1}& 2^{n-1}\\ 2^{n-1}& 2^{n-1}\end{array}\right]$
$\begin{array}{l}Ifm=3,then{A}^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right]\\ \therefore Ifm=m,then{A}^n=\left[\begin{array}{ccc}{m}^{n-1}& m^{n-1}\mathrm{........}& m^{n-1}\\ m^{n-1}& m^{n-1}\mathrm{........}& m^{n-1}\\ m^{n-1}& m^{n-1}\mathrm{........}& m^{n-1}\end{array}\right]\\ \therefore If\text{\hspace{0.17em}}{A}^n=m^{n-1}A={16}^{17}A=2^{68}A\end{array}$ 
Factors of 68 are 1,2,4,17,34,68.
If   $n-1=1\Rightarrow n=2,thenm=2^{68}$
If  $n-1=2\Rightarrow n=3,thenm=2^{34}$
If  $n-1=4\Rightarrow n=5,thenm=2^{17}$  
If  $n-1=17\Rightarrow n=18,thenm=2^4$

If  $n-1=34\Rightarrow n=35,thenm=4$
If  $n-1=68\Rightarrow n=69,thenm=2$
$\therefore$  Sum of all n=132.