HMMO

Solution:
It is straightforward to show that if $a_{1}=1$, then $a_{n}=n$ for all $n$. Since $a_{n+1}$ is an increasing function in $a_{n}$, it follows that the set of possible $a_{1}$ is of the form $\{1,2, \ldots, m\}$ for some $m$, which will be the answer to the problem.

Consider the sequence $b_{n}=a_{n+1}-1$, which has the recurrence
$$b_{n+1}=n\left\lfloor\frac{b_{n}+1}{n}\right\rfloor .$$
It has the property that $b_{n}$ is divisible by $n$. Rearranging the recurrence, we see that
$$\frac{b_{n+1}}{n+1} \leq \frac{b_{n}+1}{n+1}<\frac{b_{n+1}}{n+1}+1$$
and as the $b_{i}$ are integers, we get $b_{n+1}-1 \leq b_{n}<b_{n+1}+n$. For $n \geq 2$, this means that the largest possible value of $b_{n}$ (call this $b_{n}^{*}$ ) is the smallest multiple of $n$ which is at least $b_{n+1}$. Also, since $b_{1}=b_{0}+1$, we find $b_{0}^{*}=b_{1}^{*}-1$, meaning that the largest value for $a_{1}$ is $b_{1}^{*}$, and thus the answer is $b_{1}^{*}$. We have now derived a procedure for deriving $b_{1}^{*}$ from $b_{29}^{*}=29$. To speed up the computation, let $c_{n}=b_{n}^{*} / n$. Then, since
$$b_{n}^{*}=n\left\lceil\frac{b_{n+1}^{*}}{n}\right\rceil$$
we find
$$c_{n}=\left\lceil\frac{n+1}{n} c_{n+1}\right\rceil=c_{n+1}+\left\lceil\frac{c_{n+1}}{n}\right\rceil .$$
We now start from $c_{29}=1$ and wish to find $c_{1}$.

Applying the recurrence, we find $c_{28}=2, c_{27}=3$, and so on until we reach $c_{15}=15$. Then, $\left\lceil c_{n+1} / n\right\rceil$ becomes greater than 1 and we find $c_{14}=17, c_{13}=19$, and so on until $c_{11}=23$. The rest can be done manually, with $c_{10}=26, c_{9}=29, c_{8}=33, c_{7}=38, c_{6}=45, c_{5}=54, c_{4}=68, c_{3}=91, c_{2}=137$, and $c_{1}=274$. The last few steps may be easier to perform by converting back into the $b_{n}^{*}$.