AIME II, 2022

The graph of
$$\frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} = 1$$
is an ellipse centered at $(0, 0)$ with major axis parallel to the $x$-axis of length $2a$. The distance from the center to the foci is $\sqrt{a^2 - (a^2 - 16)} = 4$, so the foci are $F_1 = (-4, 0)$ and $F_2 = (4, 0)$.
Similarly, the graph of
$$\frac{(x - 20)^2}{b^2 - 1} + \frac{(y - 11)^2}{b^2} = 1$$
is an ellipse centered at $(20, 11)$ with major axis parallel to the $y$-axis of length $2b$. The distance from the center to the foci is $\sqrt{b^2 - (b^2 - 1)} = 1$, so the foci are $G_1 = (20, 10)$ and $G_2 = (20, 12)$.

Let $P(x, y)$ be a point that lies on both ellipses, so by a property of ellipses,
$$2a = PF_1 + PF_2 \quad \text{and} \quad 2b = PG_1 + PG_2.$$

Adding these equations and using the triangle inequality yields
$$\begin{align*} 2a + 2b &= PF_1 + PF_2 + PG_1 + PG_2 \\ &= (PF_1 + PG_1) + (PF_2 + PG_2) \\ &\geq F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} + \sqrt{16^2 + 12^2} = 46. \end{align*}$$
Thus $a + b \geq 23$. Equality is achieved by running this argument in reverse: that is, by taking $P$ to be the intersection of $\overline{F_1G_1}$ and $\overline{F_2G_2}$, and then setting $a = \frac{1}{2}(PF_1 + PF_2)$ and $b = \frac{1}{2}(PG_1 + PG_2)$. In this case, $a = 16$, $b = 7$, and $P = (14, 7.5)$. Therefore the least possible value of $a + b$ is $23$.