Let $f : \mathbb{N} \longrightarrow \mathbb{N}$ be a function such that $(x + y)f (x) \le x^2 + f (xy) + 110$, for all $x, y$ in $\mathbb{N}$. Determine the minimum and maximum values of $f (23) + f (2011)$.
1. **Define the assertion**: Let \( s(x, y) \) be the assertion of the given inequality:
   \[
   (x + y)f(x) \le x^2 + f(xy) + 110
   \]
   for all \( x, y \in \mathbb{N} \).

2. **Initial bounds on \( f(x) \)**: From the assertion \( s(x, 1) \):
   \[
   (x + 1)f(x) \le x^2 + f(x) + 110
   \]
   Rearrange to isolate \( f(x) \):
   \[
   (x + 1)f(x) - f(x) \le x^2 + 110
   \]
   \[
   xf(x) \le x^2 + 110
   \]
   \[
   f(x) \le x + \frac{110}{x}
   \]

3. **Upper bound refinement**: From the assertion \( s(x, y) \):
   \[
   (x + y)f(x) \le x^2 + f(xy) + 110
   \]
   Using the upper bound \( f(xy) \le xy + \frac{110}{xy} \):
   \[
   (x + y)f(x) \le x^2 + xy + \frac{110}{xy} + 110
   \]
   Simplify and take the limit as \( y \to \infty \):
   \[
   f(x) \le x
   \]

4. **Lower bound on \( f(x) \)**: From the assertion \( s(1, x) \):
   \[
   (1 + x)f(1) \le 1 + f(x) + 110
   \]
   Given \( f(1) = 1 \):
   \[
   1 + x \le 1 + f(x) + 110
   \]
   \[
   x \le f(x) + 110
   \]
   \[
   f(x) \ge x - 110
   \]

5. **Combining bounds**: We have:
   \[
   x - 110 \le f(x) \le x
   \]

6. **Evaluate \( f(23) \) and \( f(2011) \)**:
   \[
   23 - 110 \le f(23) \le 23 \implies 1 \le f(23) \le 23
   \]
   \[
   2011 - 110 \le f(2011) \le 2011 \implies 1901 \le f(2011) \le 2011
   \]

7. **Sum bounds**:
   \[
   f(23) + f(2011) \ge 1 + 1901 = 1902
   \]
   \[
   f(23) + f(2011) \le 23 + 2011 = 2034
   \]

8. **Conclusion**: Both bounds can be reached, for example, if \( f(x) = x \):
   \[
   f(23) + f(2011) = 23 + 2011 = 2034
   \]
   and if \( f(x) = x - 110 \) for \( x \ge 111 \):
   \[
   f(23) + f(2011) = 1 + 1901 = 1902
   \]

The final answer is \( \boxed{ 1902 \le f(23) + f(2011) \le 2034 } \)