This problem is very similar to finding the inverse of the function that is equivalent to the sum of
$$t=\sum_{n=1}^c {10 \over n}$$
The above represents the amount of time $t$ passed given which reset number $c$ the counter is on. You could ask for fractional $c$ counter reset in which case an integral or an additive term might be appropriate.
Here's a link that discusses the above. As a bonus you already intuitively know why this series diverges. Consider what happens as the number of counters increases, you don't actually expect the time to "freeze" as the number of counters reaches infinity.
Here's a link describing the harmonic numbers which are the sums of the series. You'd just find the inverse of this and get your approximate solution.
Here's a partial walk through. Assuming you know calculus, take the integral of the function inside the summation. This will roughly approximate the sum. We'll add a constant which is motivated here.
So we get,
$$t \sim 10\cdot (\ln(c)+\gamma)$$
Solving for $c$, we get
$$c \sim e^{{{t-\gamma} \over 10}}-1$$
This approximation gets better for large n, percent error wise at least.