How many ordered pairs of integers $(x, y)$ satisfy $x^{2} \leq y \leq x+6$ ?
For there to exist a pair of integers $(x, y)$ with $x^{2} \leq y \leq x+6$, it must be the case that $x^{2} \leq x+6$ and so $x^{2}-x-6 \leq 0$.

Now $x^{2}-x-6=(x-3)(x+2)$, so $x^{2}-x-6 \leq 0$ exactly when $-2 \leq x \leq 3$. (If we consider the function $f(x)=(x-3)(x+2)$, whose graph is a parabola opening upwards, its values are less than or equal to 0 between its roots.)

Therefore, any pair of integers $(x, y)$ with $x^{2} \leq y \leq x+6$ must have $x$ equal to one of $-2,-1,0,1,2,3$.

When $x=-2$, the original inequality becomes $4 \leq y \leq 4$ and so $y$ must equal 4 . There is 1 pair in this case, namely $(-2,4)$.

When $x=-1$, we obtain $1 \leq y \leq 5$ and so $y$ must equal one of $1,2,3,4,5$. There are 5 pairs in this case.

When $x=0$, we obtain $0 \leq y \leq 6$ and so $y$ must equal one of $0,1,2,3,4,5,6$. There are 7 pairs in this case.

When $x=1$, we obtain $1 \leq y \leq 7$. There are 7 pairs in this case.

When $x=2$, we obtain $4 \leq y \leq 8$. There are 5 pairs in this case.

When $x=3$, we obtain $9 \leq y \leq 9$ and so $y$ must equal 9 . There is 1 pair in this case.

In total, there are $1+5+7+7+5+1=26$ pairs of integers that satisfy the inequality.

ANSWER: 26