How many cardboard digits do I need?

Jelly, 9 bytes

‘ḶDœ|/ḟ9L

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How it works

‘ḶDœ|/ḟ9L
‘Ḷ         [0,1,...,n]
  D        convert each to list of its digits
   œ|/     fold by multiset union
      ḟ9   remove 9
        L  length

Python 2, 49 bytes

lambda n:9*len(`n`)-9+(n*9+8)/10**len(`n`)+(n<10)

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A clumsy arithmetical formula. Assume that n fits within an int so that an L isn't appended.

Thanks to Neil for saving 5 bytes by pointing out that 9's being unused could be handled by doing n*9+8 instead of n*9+9, so that, say, 999*9+8=8999 doesn't roll over to 9000.

Haskell, 117 114 108 95 89 88 87 84 82 63 bytes

6 bytes saved thanks to Laikoni

1 4 6 bytes saved thanks to nimi

g x=sum[maximum[sum[1|u<-show y,d==u]|y<-[0..x]]|d<-['0'..'8']]

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