I also don't have a reference, but I think one should be able to do this fairly explicitly as follows [it appears that Peter Taylor alludes to this strategy in their comment above]:
First, note that much easier than your original question is to count the number of tuples $(x_i,...,x_0)$ with
$$\max\{x_i,x_{i-1}-1,x_{i-2}-2,...,x_0-i\} \le i,$$ since a tuple has this property exactly when each $x_{i-j} \le j+i$. Since by assumption each $x_{i-j} \in \{1,...,N\}$, for each $j$ we have exactly $\min(i+j,N)$ choices for the value of $x_{i-j}$.
Thus we see that the number of such tuples is
$ \prod_{j=0}^i \min(i+j,N) $; in particular, if $2i \le N$, then
the number of such tuples is just $(i)(i+1)\cdots(2i-1)(2i) = \frac{(2i)!}{(i+1)!}$.
But note that we can exactly analogously compute that the
number of tuples $(x_i,...,x_0)$ with
$$ \max \{ x_i , x_{i-1} - 1,..., x_0 - i \} \le i - 1$$
is $\prod_{j=0}^i \min(i+j-1,N)$, which again is $(i-1)(i)\cdots(2i-1)$ if $2i-1 \le N$.
But the difference between these two quantities gives exactly
the number of tuples with
$$ \max \{ x_i , x_{i-1} - 1,..., x_0 - i \} = i,$$
which is what you were looking for. So, for instance if $2i \le N$, the quantity you're looking for is
$$ i(i+1)\cdots(2i) - (i-1)(i)\cdots(2i-1)
= [(2i - (i-1))](i)(i+1)\cdots(2i-1) = (i+1) \frac{(2i-1)!}{(i-1)!}. $$
The more general lesson here is that the condition $\max(a bunch of junk)=c$ is often hard to get your hands on; but the condition $max(a bunch of junk) \le c$ is equivalent to $j \le c$ for each piece of junk $j$, which is often easier. Thus, its often fruitful to convert problems of the former type into problem of the latter type, just as we did here.
