We can go from one to sixty quickly, right?

It can be shown that $N$ has 111 digits in total; thus, $M$ has 11 digits. Since 9 only appears 6 times between 1 and 60, we first make 9 the 5 leftmost digits of $m$ , and seek to maximise the 6 rightmost digits. We only need to look at which digits among 505152...5960 we have to delete. Considering that we must end up with 6 digits, it is easy to see that they are 785960, and so $M = \boxed{99999785960}$ .

The answer is 99999785960. We can't start with all six 9s, since that would leave us with only 2 more digits, whereas the answer has to have 11 digits. So we look for an answer starting with 5 nines. The sixth digit can't be 8, since that would leave us with only 4 more digits after 999998, whereas we need 5. So we look for an answer starting with 999997. The remaining digits to choose from are 585960, and we need to select five from among them: we therefore omit the first digit that is less than its successor, which is 5, leaving us with the five digits 85960. The answer is therefore 99999785960.

What an amazing question.