Problem 589

5.61.

Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the distances from the center $O$ of the ellipse to the endpoints of the major axis, the endpoints of the minor axis, the foci, and the directrices are $a$, $b$, $c$, $d$ respectively, and satisfy
$$c^{2}=a^{2}-b^{2}, d=\frac{a^{2}}{c} \text {. }$$

Since $a$, $b$, $c$ form a Pythagorean triple, the Pythagorean triples satisfying $a \leqslant 20$ are
$$\begin{aligned} \{a, b, c\}= & \{3,4,5\},\{6,8,10\},\{9,12,15\}, \\ & \{12,16,20\},\{5,12,13\},\{8,15,17\}, \end{aligned}$$

Among them, only $\frac{15^{2}}{9}=25$ and $\frac{20^{2}}{16}=25$.
Upon verification, when $(a, b, c, d)=(15,12,9,25)$, the value of $a+b+c+d$ is the smallest.
At this time, $a+b+c+d=61$.