I played the following game (having in mind that if one wants to maximize, then take a $10$ -digit number):

$0 0 0 0 0 0 0 0 0 0$

One starts putting something in the $9$ th position. Logically, we only can put $1$ . Otherwise, we would get a number with more than $10$ digits. The same logic applies for the $8$ th , $7$ th and $6$ th positions. So we put $1$ and we try whatever would imply putting that $1$ .

For example, putting $1$ on the $9$ th position implies having a $9$ on the $0$ th position, and then the number doesn't works because we would get $8$ zeroes, which contradicts having a $9$ on the $0$ th position. So we pass to the eight position, and seventh position, and similar things will occur (deserves long explanation, but is the same idea... sorry im lazy to write). When we get to the $6$ th position, we are able to put a $6$ on the $0$ th position, while we have $8$ zeroes. On the $1$ th position we put a $2$ , and on the $2$ th position a $1$ . That is:

$6 2 1 0 0 0 1 0 0 0$

So we've maximized... the answer: $\boxed {6210001000}$ .

We have that to maximize $n$ , so $n$ have 10 digits. Be $n=\overline{a_1 a_2 a_3 \cdots a_{10}}$ , is autodescriptive so $a_1+ \cdots + a_{10}=10$ . If $a_1=9$ , then $a_2+\cdots +a_{10}=1$ , but is not posible, because $a_{10}=1$ and $a_2=1$ . Contradiction. If $a_1=8$ , then $a_2+\cdots +a_{10}=2$ , but is not posible, because $a_9=1$ and $a_2=2$ and $a_2=1$ . Contradiction. If $a_1=7$ , then $a_2+\cdots +a_{10}=3$ , but is not posible, because $a_8=1$ and $a_2=2$ and $a_2=1$ . Contradiction. If $a_1=6$ , then $a_2+\cdots +a_{10}=4$ , is posible, when $n=6210001000$ .

We want to number to be as large as possible. First, we try to place $9$ at the most significant digit.

9000000000
0123456789

This can't work because there is $1$ nine and there have to be $9$ zero's for the number to be self descriptive.

Then, we try $8$ .

8200000010 
0123456789

This only has $7$ zeros but we need $8$ .

After iterating a couple of times, we find a suitable candidate at a number that starts with $6$ . Now, this number has to be the greatest because we tried and failed for number that start with $9,8,7$ . Any other combination that we try will be $< 6210001000$ . Thus, finally we get $\boxed{6210001000}$ .

Note two things first -

i) It can not have more than ten digits. ii) It should have maximum zeroes.

Step 1. Let's try nine zeroes

9000000000

But it should be 9000000001, because there is one '9' (yes, this is also wrong but we have to get the first digit first). But now there are 8 zeroes.

Step 2. Now let's try eight zeroes.

8000000010

But it should be 8100000010, because there is one '1'. But now there are 7 zeroes.

Step 3. Now let's try seven zeroes

7100000100

But it should be 7210001000, because there are two '1's and one '2'. But now there are 6 zeroes.

Step 4. Let's try Six zeroes, then.

6210001000

Yep, that's the answer.

ZiSongYeoh

But the answer is not the largest self-descriptive number. You should think on these self-descriptive numbers, 72100001000, 821000001000 and so on, What do you say?

You should clarify that the number can have a maximum of 10 digits. One could interpret the problem as saying that (for example) the 11th digit (digit number 10), which indicates how many times the digit "10" appears in the number, must equal the number of times the two-digit string "10" occurs, or that it must equal zero because "10" is not a digit, so of course the digit "10" appears zero times. Using the latter interpretation (which is the interpretation I first assumed) the answer would be 9210000001000. I haven't tried to solve the problem with the former interpretation.