An inclusion-exclusion argument gives us the following for the distributions of the $A_i$.
We first consider the case of the $n$ numbers being distinct (selection without replacement).
If $1\leqslant n \leqslant m$, $1\leqslant i \leqslant m+1-n$ and $0\leqslant k \leqslant M_i$, then we have:
$ \textrm{P}(A_i=k)\ =\ P\:(m,n,i,k)\ =\ \frac{1}{T}\: \sum_{j=k}^{M_i} (-1)^{j-k} {j\choose k} {n\choose j} {m-ij\choose n-j} $
where $T={m\choose n}$ is the total number of configurations, and $M_i=\textrm{min}(n,\lfloor\frac{m-n}{i-1}\rfloor)$ is the maximum possible gap size.
$Q_k={n\choose k} {m-ik\choose n-k}$ is a count of the combinations with at least $k$ gaps of size $i$, determined by first choosing $k$ gaps of size $i$ and then arbitrarily choosing how the remaining 'space' is split up. However $Q_k$ counts combinations with more than $k$ $i$-gaps multiple times; indeed combinations with $j$ $i$-gaps are counted $j \choose k$ times. Thus if $R_k$ is the number of combinations with exactly $k$ gaps of size $i$, we have $R_k=Q_k-\sum_{j=k+1}^{M_i}{j \choose k}R_j$. Inclusion-exclusion gives us that $R_k=T\times P\:(m,n,i,k)$.
If we now consider the case of the $n$ numbers not necessarily being distinct (selection with replacement), then, if we include $0$ in the set of numbers that may be selected (so that the first gap can be $0$ like the others), we have:
$ \textrm{P}(A_i=k)\ =\ P\:(m+n,n,i+1,k) $
where $P\:$ is as above.
This follows simply from the fact that selecting $n$ numbers from $m+1$ with replacement is equivalent to selecting $n$ numbers from $m+n$ without replacement, but with each gap being smaller in size by $1$.