A palindrome with an even number of digits has each digit appearing in a mirrored pair, so its digit sum is even. Since $13$ is odd, the palindrome has an odd number of digits, and if $m$ is the middle digit, the rest of the digit sum $13 - m$ is split evenly between the two halves, so $m$ is odd. A one-digit palindrome would need $m = 13,$ which is impossible.

The palindrome is determined by its middle digit $m$ and the block of digits to the left of center: a nonempty string of nonzero digits with sum $s = \frac{13 - m}{2}.$ For $m = 1, 3, 5, 7, 9$ we get $s = 6, 5, 4, 3, 2.$ Since $s \le 6,$ every digit of such a string is automatically at most $9,$ so the number of strings is the number of compositions of $s,$ which is $2^{s-1}$ (each of the $s - 1$ gaps between units is either a break or not).

The total is $$\begin{aligned} &2^{5} + 2^{4} + 2^{3} + 2^{2} + 2^{1} \\ &= 32 + 16 + 8 + 4 + 2 \\ &= 62. \end{aligned}$$