You have a discretized version of the negative log distribution, that is, the distribution whose support is $[0, 1]$ and whose pdf is $f(t) = - \log t$.
To see this, I'm going to redefine your random variable to take values in the set $\{ 0, 1/N, 2/N, \ldots, 1 \}$ instead of $\{0, 1, 2, \ldots, N \}$ and call the resulting distribution $T$. Then, my claim is that
$$ Pr\left( T = \frac{t}{N} \right) \rightarrow - \frac{1}{N} \log \left( \frac{t}{N} \right) $$
as $N, t \rightarrow \infty$ while $\frac{t}{N}$ is held (approximately) constant.
First, a little simulation experiment demonstrating this convergence. Here's a small implementation of a sampler from your distribution:
t_sample <- function(N, size) {
bounds <- sample(1:N, size=size, replace=TRUE)
samples <- sapply(bounds, function(t) {sample(1:t, size=1)})
samples / N
}
Here's a histogram of a large sample taken from your distribution:
ss <- t_sample(100, 200000)
hist(ss, freq=FALSE, breaks=50)

and here's the logarithmic pdf overlaid:
linsp <- 1:100 / 100
lines(linsp, -log(linsp))

To see why this convergence occurs, start with your expression
$$ Pr \left( T = \frac{t}{N} \right) = \frac{1}{N} \sum_{j=t}^N \frac{1}{j} $$
and multiply and divide by $N$
$$ Pr \left( T = \frac{t}{N} \right) = \frac{1}{N} \sum_{j=t}^N \frac{N}{j} \frac{1}{N} $$
The summation is now a Riemann sum for the function $g(x) = \frac{1}{x}$, integrated from $\frac{t}{N}$ to $1$. That is, for large $N$,
$$ Pr \left( T = \frac{t}{N} \right) \approx \frac{1}{N} \int_{\frac{t}{N}}^1 \frac{1}{x} dx = - \frac{1}{N} \log \left( \frac{t}{N} \right)$$
which is the expression I wanted to arrive at.