Determine the solutions

Multiply by $y$ and subtract $x$ to get $\frac{x^2y^2+xy+y^2}{xy^2+y+11} - x = \frac{y^2-11x}{xy^2+y+11}.$ This must also be an integer. If $y^2-11x$ is positive, this is impossible, because the numerator will be less than the denominator. If $y^2=11x,$ then $y=11k$ for some integer $k$ and $x=11k^2.$ The tuple $(11k^2,11k)$ gives a solution for any positive integer $k.$

Finally, if $y^2 < 11x,$ then we need $11x-y^2$ to be at least as large as the denominator: $11x-y^2 \ge xy^2 + y + 11 \Rightarrow x(y^2-11)+(y^2+y+11) \le 0.$ Note that if $y \ge 4,$ the right side equals $m+nx$ where $m$ and $n$ are positive, which would be a contradiction.

So this leaves the cases $y=1,2,3.$

If $y=1$ we get $\frac{x^2+x+1}{x+12}$ is an integer, but this equals $x-11 + \frac{133}{x+12},$ and the only divisors of $133$ larger than $12$ are $19$ and $133,$ so the two solutions are $x=7,121.$

If $y=2$ we get $\frac{2x^2+x+2}{4x+13}$ is an integer, but this equals $\frac12 \left( x - \frac{11x-4}{4x+13}\right).$ So $\frac{11x-4}{4x+13}$ is an integer. Clearly this integer is less than $3,$ so it's $1$ or $2.$ If it's $1$ then $7x=17,$ which is no good. If it's $2$ then $x=10,$ and that does in fact work (the expression is $\frac{212}{53}=4$ ), so that's the one solution in this case.

If $y=3$ then we get $\frac{3x^2+x+3}{9x+14}$ is an integer, but this equals $\frac13 \left( x - \frac{11x-9}{9x+14} \right).$ So $\frac{11x-9}{9x+14}$ is an integer. Clearly this integer is less than $2,$ so it's $1.$ But this leads to $2x=23,$ which is no good.

Putting it all together, our solutions are $(7,1),(121,1),(10,2),$ and $(11k^2,11k)$ for any $k.$ The sum is $7+1+121+1+10+2+11+11=\fbox{164}.$