Let $a = 110$. We will first show that a necessary and sufficient condition for a positive integer-valued function $f$ defined on the set positive integers to satisfy the given inequality is that $f$ satisfies the following simpler inequality:
$$(\dagger) \quad t - a \le f(t) \le t \quad \text{for any positive integer } t.$$
To see this, first note that for any positive integer $s$, substituting $(x, y) = (s, 1)$ into the given inequality, we obtain $(s+1)f(s) \le s^2 + f(s) + a$, from which we get $f(s) \le s + \frac{a}{s}$. Next, for any positive integer $t$, substitute $(x, y) = (t, 2a)$ into the given inequality and using the fact $f(2at) \le 2at + \frac{1}{2t}$ which follows from the result above, we get
$$(t + 2a)f(t) \le t^2 + f(2at) + a \le t^2 + 2at + \frac{1}{2t} + a,$$
which yields further that
$$f(t) \le t + \frac{1}{2t(t + 2a)} + \frac{a}{t + 2a} < t + \frac{1}{2} + \frac{1}{2} = t + 1.$$
Since $f$ is positive integer-valued, we conclude that $1 \le f(t) \le t$, and in particular, $f(1) = 1$. Finally, substituting $(x, y) = (1, t)$ into the given inequality, we get $(1+t) \cdot 1 \le 1 + f(t) + a$, from which we get $t - a \le f(t)$, and thus we have shown that the function $f$ satisfying the given inequality satisfies the inequality $(\dagger)$ for any positive integer $t$.
Conversely, suppose a positive integer-valued function $f$ satisfies the inequality $(\dagger)$ for any positive integer $t$. Then, for any pair of positive integers $x, y$, we have
$$(x + y)f(x) \le (x + y)x = x^2 + (xy - a) + a \le x^2 + f(xy) + a,$$
so that $f$ satisfies the inequality given for the problem.
Thus, if $f$ satisfies the given inequality, then we have
$$1 \le f(23) \le 23, \quad 1901 = 2011 - 110 \le f(2011) \le 2011,$$
and since the value of $f(23)$ and $f(2011)$ can be chosen to take any of the integer values in $[1, 23]$ and $[1901, 2011]$, we conclude that the minimum possible value for $f(23) + f(2011)$ is $1 + 1901 = 1902$ and the maximum possible value is $23 + 2011 = 2034$.