Both equations are ellipses through a common point: the first has foci $(\pm 4, 0),$ the second has foci $(20, 10)$ and $(20, 12)$

$2a + 2b$ is the sum of the four focal distances of the common point; pair the foci so the triangle inequality gives segments that actually intersect

The first ellipse has $c^2 = a^2 - (a^2 - 16) = 16,$ hence foci $F_1 = (-4, 0)$ and $F_2 = (4, 0),$ with distance sum $2a.$ The second is centered at $(20, 11)$ with vertical major axis and $c^2 = b^2 - (b^2 - 1) = 1,$ hence foci $G_1 = (20, 10)$ and $G_2 = (20, 12),$ with distance sum $2b.$ If $P = (x, y)$ lies on both, then $$\begin{aligned} 2a + 2b \\ &= (PF_1 + PG_1) \\ &\quad {}+ (PF_2 + PG_2) \\ &\ge F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} \\ &\quad {}+ \sqrt{16^2 + 12^2} \\ &= 26 + 20 \\ &= 46, \end{aligned}$$ so $a + b \ge 23.$

Equality requires $P$ to lie on both segments $\overline{F_1G_1}$ and $\overline{F_2G_2}.$ These segments do intersect, at $P = \left(14, \frac{15}{2}\right):$ then $PF_1 = \frac{39}{2},$ $PF_2 = \frac{25}{2},$ so $2a = 32,$ $a = 16 \gt 4;$ and $PG_1 = \frac{13}{2},$ $PG_2 = \frac{15}{2},$ so $2b = 14,$ $b = 7 \gt 1.$

Hence the least possible value of $a + b$ is $16 + 7 = 23.$