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Let $N = \overline{abcde}$ be a 5-digit palindrome, where $1\leq a \leq 9$ , $0\leq b,c,d,e\leq 9$ . For $N$ to be a palindrome, we must have $a = e$ and $b = d$ , so $N = \overline{abcba}$ . Thus by the rule of product, there are $9 \times 10 \times 10 \times 1 \times 1 = 900$ 5-digit palindromes.

For $N=\overline{abcba}$ to be divisible by $4$ the last two digits must be divisible by $4$ . There are $25$ possible values for the last two digits (i.e. $00, 04, 08, 12, \ldots 92, 96$ ). However any number that ends in $0$ must be excluded since $N$ is a 5-digit palindrome and the first digit cannot be $0$ . $5$ of the $25$ numbers end in $0$ (i.e. $00, 20, 40, 60, 80$ ). Thus there are $25 - 5 = 20$ possibilities for the last two digits. The middle digit can be any digit from $0$ to $9$ and the first two digits are fixed by the last two digits. Therefore, by the rule of the product there are $20 \times 10 = 200$ 5-digit palindromes that are divisible by $4$ .

Thus the probability is $\frac{200}{900} = \frac{2}{9}$ . Hence $a + b = 2 + 9 = 11$ .