Before beginning, note that in questions like these, we prefer to say that leading zeroes are not written. That is to say that we do not consider $00500$ a palindrome number since it could be reduced to $500$, ignoring the zeroes on the left.
Break this into cases: it is 1 digit long, it is 2 digits long, it is 3 digits long, $\cdots$, it is 9 digits long, (and it is 10 digits long)
In the case of it being five digits long we have:
__, __, __, __, __
The furthest left digit must match the furthest right, the second on the left must match the second on the right, and the one in the middle could be anything.
By the multiplication principle, http://en.wikipedia.org/wiki/Rule_of_product, you can break this into steps. Step 1: Choose the first digit (any digit 1-9, in doing so, that forces the last digit to be the same), Step 2: Choose the second digit (any digit 0-9, in doing so, that forces the second to last digit to be the same), Step 3: Choose the digit in the middle (any digit 0-9).
Note also that in step 1, the first digit is not allowed to be a zero, but in all other cases it is allowed.
So, we have for the case of a 5 digit palindrome number, $9\cdot 10 \cdot 10\cdot 1 \cdot 1$ number of possibilities.
Repeat this process for all other cases and add them together for the total amount.
Note also for the case of it being 10 digits, you will also need to check that it is less than 1,000,000,000. You should quickly notice that since 1,000,000,000 is the smallest 10 digit number, and is not itself a palindrome, all other 10 digit numbers are not smaller than 1,000,000,000 and thus are ignored for the purposes of this problem.
Hint: If you pick any four distinct numbers from the set $\{0,1,2,3,4,5,6,7,8,9\}$, there is one and only one way to arrange them so that they are in decreasing order.
For the numbers whose digits are in increasing order, simply observe that the leftmost (most significant, leading) digit cannot be zero.
Best Answer
Details depend on whether for example $0110$ counts as a $4$-digit palindrome. We will suppose it doesn't. This makes things a little harder.
If $n$ is even, say $n=2m$, the first digit can be any of $9$, then the next $m-1$ can be any of $10$, and then the rest are determined. So there are $9\cdot 10^{m-1}$ palindromes with $2m$ digits.
If $n$ is odd, say $n=2m+1$, then the same sort of reasoning yields the answer $9\cdot 10^{m}$.