Are smaller divisors more "likely" to produce integer and near-integer quotients? For example, if we divide 100 by 50, then 100 by 49.8, then 49.8, etc. down to 100 divided by 1, we will have a list of 491 quotients, 10 of which are integers (2, 4, 5, 8, 10, 20, 25, 40, 50, 100). For the first 250 divisors (50.0 through 25.1), there is only one integer quotient (2). For the last 41 divisors (5.0 through 1), there are five integer quotients (20, 25, 40, 50, 100). So it seems that the smaller the divisor, the more "likely" it is to produce an integer quotient.
But are smaller divisors also more "likely" to produce near-integer quotients, defined as being within 0.1 of an integer? Using the same example as above, for the first 250 divisors, only one quotient is an integer, but 51 are near-integer. For the last 241 quotients, nine are integers but only 39 are near-integers.
So it seems that the answer is "yes" for integer quotients, but "no" for non-integer quotients. But I would love to know for sure, i.e. how to write a proof for this.
Incidentally, I need to know this for my research on time series data in a biological system.
 A: Think about what you’re asking graphically. You are looking for values of $100\over d$ that are integers (or near-integers), where $d$ is a multiple of $0.1$ between $1$ and $50$. Here’s a partial graph of the points $(d,{100\over d})$, with horizontal lines through the near-integer values of ${100\over d}$.

Here's the same graph with exact integer values of $100\over d$ marked.

And here’s the first (near-integers) graph, but using more $d$-values (multiples of $0.05$).

As the graphs suggest, once $100\over d$ gets small, the number of $d$ values between successive integers increases. But whether you hit or get near every integer along the way depends on a) how dense your set of divisors is, b) those divisors’ potential commensurability with your dividend (especially if you are seeking integers, not near-integers), and c) your criterion for “near.”
I’ve shown only the dividend $100$, divisors spaced by $0.1$ and $0.05$ (both of which will include some exact divisors of $100$), and a “nearness” criterion of within-$0.1$. Between these graphs and Carl’s answer, you can get a feel for how each consideration affects the outcome.
A: Take a real number $r$. You're asking how likely it is that, for a given integer $n$, $n/r$ is within some tolerance $\epsilon$ (in your question, it's $0.1$) of an integer.
Imagine moving along the interval $[0,1)$, starting at $0$. Every second, you add $1/r$. If you ever reach a number greater than $1$, you'll subtract $1$ so that you're back in the interval $[0,1)$. Intuitively, we should expect that the numbers we reach are uniformly distributed within this interval, in that there shouldn't be any reason that any number is more likely than any other number to show up.
If $r$ is a rational $a/b$ (with $\gcd(a,b)=1$), then the numbers we get are exactly $bn/a$, and they should be uniformly distributed among the rational numbers with denominator $a$. So, the probability that our result is within $\epsilon$ of an integer is $1/a$ times the number of integers $t$ in the range $[0,a)$ so that $t/a\leq \epsilon$ or $t/a>1-\epsilon$. This is just
$$\frac{2\lfloor a\epsilon\rfloor+1}{a}$$
(the $+1$ comes from $t=0$). As $a$ grows large, this tends to $2\epsilon$. If $\epsilon=0$, i.e. we require it to be an integer instead of a near-integer, it is $1/a$, which tends to $0$ as $a\to\infty$. This explains some of the phenomena you saw.
Side-note: if $r$ is irrational, then the results are uniformly distributed by the equidistribution theorem, and the answer is actually $2\epsilon$ for near-integer and $0$ for integer.
A: Using $0.1$ as your step value and $q$ as your quotient is the exact same thing as using $1$ as your step value and $10q$ as your quotient.
And the number of integers you get will depend ENTIRELY on how many divisors the quotient has.
So the number of integers generated but $\frac {q}{n\cdot r}$ will have is determined by how many divisors $\frac 1r q$ has.  ANd the smaller $r$ is the larger $\frac 1r q$ is and the larger a number is the more divisors it will likely have.
I can tweak and exception.  $\frac {24}n$ will have $8$ integers because $24$ has $8$ divisors.  $\frac {24}{0.1n}$ will have $32$ integers because $240$ has $32$ divisors.  But $\frac{24}{\frac {24}{1151}}$ will yield $2$ integers because $24\times \frac 1{\frac {24}{1151} } =1151$ which is prime and only has $2$ divisors.
