The maximum value can be found easily: Putting y=1 we get $(x+1)f(x)\le x^2+f(x)+110$ so $xf(x)\le x^2+110$ which leads to$ f(x)\le x+\frac{110}{x}$ Now just putting x=23 and x=2011 we get the result $f(23)+f(2011)\le 23+2011+110(\frac{1}{23}+\frac{1}{2011})$ Therefore $max(f(23)+f(2011))=2038.837$ The minimum value is still yet to be done.

from where is this problem??

It is one of the Indian olympiad sums.

It is a postal coaching problem Let $s(x,y)$ be the assertion $(x+y)f(x)\le x^2+f(xy)+110$, any solution is such that $x-110\le f(x)\le x$ $\forall x$ $s(x,1)$ $\implies$ $f(x)\le x+\frac{110}x$ Then $s(x,y)$ $\implies$ $(x+y)f(x)\le x^2+xy+\frac{110}{xy}+110$ and so $f(x)\le x+110\frac{xy+1}{xy(x+y)}$ and so, setting $y\to +\infty$ : $f(x)\le x$ $\forall x$ So $f(1)=1$ $s(1,x)$ $\implies$ $f(x)\ge x-110$ any function from $\mathbb N\to\mathbb N$ such that $x-110\le f(x)\le x$ $\forall x$ is a solution $(x+y)f(x)\le (x+y)x=x^2+xy\le x^2+f(xy)+110$ $f(23)\le 23$ and $f(2011)\le 2011$ and so $f(23)+f(2011)\le 2034$ $f(x)=x$ is a solution and so $f(23)+f(2011)= 2034$ may be reached. $f(23)\ge 1$ and $f(2011)\ge 2011-110= 1901$ and so $f(23)+f(2011)\ge 1902$ so $f(23)+f(2011)= 1902$ may be reached. Hence the answer ${1902\le f(23)+f(2011)\le 2034}$ and both bounds may be reached.