This is a community wiki answer to summarize the main results we've got for quick access. Feel free to edit and add more results. Theoretical achievements are here:
Result. $3\mid n$, by virtually everyone.
Result. $n\equiv 1\pmod 2$, and, thus, $n\equiv 3\pmod 6$, obtained by benh. See also the message on chat.
Result. if a prime $p$ divides $n^2-1$, then $p\equiv 2^k\pmod{15}$ for some $k$, obtained by benh.
Result. $n^2-1$ isn't divisible by $3, 5, 7, 11$. Obtained by Yiorgos S. Smyrlis.
Result. $n\equiv \pm 3\pmod 8$. Obtained by Tim Ratigan.
Result. combining the congruences, $n\equiv 3, 93 \pmod{120}$. See a proof here.
Result. Jack D'Aurizio was able to rule out $n\not\equiv \pm1\pmod p$ for every prime number $p>5$ for which $(3\cdot 5^{-1})$ has an odd order $\pmod{p}$ or an order divisible by $4$, see here. In combination with benh's result this gives that the smallest odd prime factor of $n^2-1$ is at least $19$.
Numerical checkings are given and updated in this part.
Result. Listing was able to verify by brute force that $n=3 \lor n>10^{12}$, extending the result by Tapio Rajala that $n=3 \lor n>10^{11}$.
Add your own result or someone else's. Please give proper credit and don't post the proofs here; link them instead. For further discussion, e.g. disproving or strengthening any claim, use this chatroom.
This turned out to be a long standing open problem. Needless to say, breakthroughs in this question will be very well rewarded. I don't want this question to stop here, so I'll offer a +100 bounty very soon. Keep the good work up!
Excellent question! The answer is $127$ and $128$... but why? If you wanted to find a number divisible by $1,2,3,4$ you might first multiply these numbers and say $24$. However, you soon realize $4$ is already a multiple of $2$; you can use just $3\times4$ to get $12$. Therefore, you need only multiply the largest powers of the primes that factor all of the digits from $2$ to $200$ to get a number that is divisible by all of the integers from $1$ to $200$.
If you do this; you will find the number is $2^7\cdot3^4\cdot5^3\cdot7^2\cdot11^2\cdot13^2\cdot17\cdot19\cdot23\cdot29\cdot\ldots$(the rest of the primes up to $199$) = a very large number.
Next we need to find a restriction to eliminate two consecutive numbers. One of the two numbers must be even. The only way to remove an even number from the above calculation without modifying any of the other primes is to reduce the power of $2^7$ to $2^6$; this removes the number $128$ from the list. Since $127$ is also a prime number, it can also be removed from the list without affecting any of the other primes in the list...
I hope this helps.
Best Answer
For four or more digit numbers, we do have a divisibility test by $13$ owing to $1001$ being divisible by $13$ ie, $1000$ leaving remainder $-1$. One checks the alternating sum of blocks of three consecutive digits. Eg, $\overline{abcdefgh}$ is divisible by $13$ if $\overline{ab}-\overline{cde}+\overline{fgh}$ is.
So we check how many $n$ there for which following is multiple of $13$ $$\overline{3}-\overline{0a0}+\overline{b03}=\overline{(b-1)(10-a)6}$$
We have $130+26=156$ one such number $\Rightarrow b=2$, $a=5$.
Others can be found by adding/subtracting $130$.