One Ring to rule them all. One String to contain them all
GolfScript (35 31 26 chars)
10,{:x),{:&x=x+,{x&@}/}/}/
Output is
000100110111200201210211220221222300301302310311312320321322330331332333400401402403410411412413420421422423430431432433440441442443444500501502503504510511512513514520521522523524530531532533534540541542543544550551552553554555600601602603604605610611612613614615620621622623624625630631632633634635640641642643644645650651652653654655660661662663664665666700701702703704705706710711712713714715716720721722723724725726730731732733734735736740741742743744745746750751752753754755756760761762763764765766770771772773774775776777800801802803804805806807810811812813814815816817820821822823824825826827830831832833834835836837840841842843844845846847850851852853854855856857860861862863864865866867870871872873874875876877880881882883884885886887888900901902903904905906907908910911912913914915916917918920921922923924925926927928930931932933934935936937938940941942943944945946947948950951952953954955956957958960961962963964965966967968970971972973974975976977978980981982983984985986987988990991992993994995996997998999
(1020 chars) This is a variant on the Lyndon word concatenation approach: rather than use the primitive 1-char words it uses multiples of 111 for shorter code but repeated occurrences of those numbers; and rather than use minimal elements of the conjugacy groups it uses maximal elements, because that shortens the loops.
10,:^{:x^>{x.@:&<+^>{.x>{x&@}*}/}/}%3>0.
at 40 chars (can probably still be improved) generates an optimal string, which is of length 999 chars:
100200300400500600700800901101201301401501601701801902102202302402502602702802903103203303403503603703803904104204304404504604704804905105205305405505605705805906106206306406506606706806907107207307407507607707807908108208308408508608708808909109209309409509609709809911121131141151161171181191221231241251261271281291321331341351361371381391421431441451461471481491521531541551561571581591621631641651661671681691721731741751761771781791821831841851861871881891921931941951961971981992223224225226227228229233234235236237238239243244245246247248249253254255256257258259263264265266267268269273274275276277278279283284285286287288289293294295296297298299333433533633733833934434534634734834935435535635735835936436536636736836937437537637737837938438538638738838939439539639739839944454464474484494554564574584594654664674684694754764774784794854864874884894954964974984995556557558559566567568569576577578579586587588589596597598599666766866967767867968768868969769869977787797887897987998889899900
Trying to make this do reverse strings runs into problems with omitting the multiples of 111.
To see that 999 is the optimal length (since my brief comments above don't convince everyone), start from the full de Bruijn sequence which (taken as a cyclic string) contains every 3-digit sequence of characters from 0 to 9. Since there are 1000 of them, it must be at least 1000 characters long; that it can be precisely 1000 characters long is usually proven by an Eulerian walk on a graph whose nodes are two-digit sequences xy with 10 edges, each labelled with one digit z, which take xy to yz.
We don't need sequences beginning 0, so given a de Bruijn sequence we can rotate to put 000 at the end. Then we don't need either of the sequences which wrap round to the beginning, but we do need two of the 0s to finish the sequence starting with the digit before 000, so we can delete one of them to get a 999-character string. Every remaining 0 is used in a number which doesn't begin with 0.
GolfScript, 17 characters
999,{`1$1$?0<*+}/
Plain approach to add each number if not already present in the string (note: 999 is not checked or added, but contained already in the output).
Output is 1133 characters:
01234567891011131415161718192021222425262728293032333536373839404344464748495054555758596065666869707677798087889099100102103104105106107108109110112114115116117118119120124125126127128129130132133134135136137138139140142143144145146147148149150152153154155156157158159160162163164165166167168169170172173174175176177178179180182183184185186187188189190193194195196197198199200203204205206207208209219220221223225226227228229230231235236237238239240243244245246247248249250253254255256257258259260263264265266267268269270273274275276277278279280283284285286287288289290294295296297298299300304305306307308309311329330332334336337338339340342346347348349350354355356357358359360364365366367368369370374375376377378379380384385386387388389390395396397398399400405406407408409422439440443445447448449450453457458459460465466467468469470475476477478479480485486487488489490496497498499500506507508509533549550554556558559560564568569570576577578579580586587588589590597598599600607608609644659660665667669670675679680687688689690698699700708709755769770776778780786790797799800809866877879880887888897898899900908932943954965976979987989
I don't have any code, but I thought someone might appreciate this intuitive proof that 999 characters is the lower bound to the length of the output:
First, every 1- and 2-digit number is part of a 3-digit number, so ignore everything less than 100. 100-999 inclusive is 900 3-digit numbers.
The most optimal way to solve the problem is if every character is used as much as possible. That means the numbers overlap as much as possible, like this:
123
234
345
The first number will therefore add 3 characters, and each subsequent number will add 1 character. That gives 3 + 899 = 902 characters as a lower bound.
However, when there is a zero, we can't use it to start a new 3-digit number. We can reuse it in the middle of another 3-digit number though, as long as it is not followed by another zero:
120
203 <- Ok.
034 <- not a number 100-999.
But:
100
002 <- not a number 100-999.
023 <- not a number 100-999.
Therefore, each zero which appears in the output extends the output by 1 character - except for the last two characters which may be zero as they do not overlap any further numbers:
???
??0
?00
There are 81 numbers with strictly one zero in the middle (?0?), 81 with strictly one zero at the end (??0), and 9 with two zeros (?00).
Every ??0 number can share a zero with either a ?0? number or a ?00 number, but not both. ?0? and ?00 can never share zeros, so there must be at least 81 + 9*2 zeros in the output.
This gives a lower bound of 3 + 899 + 81 + 9*2 - 2 = 999 characters.
Apologies if this is considered off-topic, but it was too long to fit in a comment.