Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$. Then \[ 100a + 10b+c = 11m\] for some $m$. We also have $m=a^2+b^2+c^2$. Substituting this into the first equation and simplifying the result yield \[100a+10b+c = 11a^2 +11b^2 +11c^2\] For an integer divisible by $11$, the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by $11$. Thus we get: $$b = a + c\quad\text{or}\quad b = a + c - 11$$ Case $1$: Let $b=a+c$. We get \[100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2\]\[10a+c = 2a^2+2ac+2c^2\] Since the right side is even, the left side must also be even. Let $c=2q$ for some $q = 0,1,2,3,4$. Then \[10a+2q=2a^2+4aq+8q^2\]\[5a+q=a^2+2aq+3q^2\] Substitute $q=0,1,2,3,4$ into the last equation and then solve for $a$. When $q=0$, we get $a=5$. Thus $c=0$ and $b=5$. We get that $N=550$ which works. When $q=1$, we get that $a$ is not an integer. There is no $N$ for this case. When $q=2$, we get that $a$ is not an integer. There is no $N$ for this case. When $q=3$, we get that $a$ is not an integer. There is no $N$ for this case. When $q=4$, we get that $a$ is not an integer. There is no $N$ for this case. Case $2$: Let $b = a + c - 11$. We get \[100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)\]\[ 10a+c=2a^2+2c^2+2ac-22a-22c+131\]\[2(a-8)^2+2(c-\frac{23}{4})^2+2ac-\frac{505}{8}=0\] Now we test all $c=0\rightarrow10$. When $c=0,1,2,4,5,6,7,8,9$, we get no integer solution to $a$. Thus, for these values of $c$, there is no valid $N$. However, when $c=3$, we get \[2(a-8)^2+2(3-\frac{23}{4})^2+6a-\frac{505}{8}=0\]\[2(a-8)^2+6a-48 = 0\] We get that $a=8$ is a valid solution. For this case, we get $a=8,b=0,c=3$, so $N=803$, and this is a valid value. Thus, the answers are $\boxed{N=550,803}$