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First,we need to understand that the expression will be maximum only when all $x_i$ are positive. Now,

$\displaystyle{\sum_{i=1}^n i x_{i}} \leq {(\displaystyle{\sum_{i=1}^n i ^2})}^{\frac {1}{2}}*(\displaystyle{\sum_{i=1}^n x_{i}^{2}})^{\frac {1}{2}}$

Thus, $M_p = (\displaystyle{\sum_{i=1}^p i^{2}})^{\frac {1}{2}}$

Since, $\displaystyle{\sum_{i=1}^n x_{i}^{2}} = 1$

Now, we need to find integer values of $M_p$ ,which is equal to

$(\frac {p(p+1)(2p+1)}{6})^{\frac {1}{2}}$

Now,for $M_p$ to be integer,the entire expression should inside root must be a perfect square.I had to use hit and trial in order to find the values.

they are perfect squares for $p=1$ and $p=24$ .

Thus,Ans=1+24=25

However,Please let me know if you are able to find a method to find values of p without hit and trial.