Specific robust measure of scale Do you know of any outlier resistant measure of scale $R(Y_1, \dots, Y_n)$ with the following property related to the sample standard deviation $S$?
Magic property:
If $S(Y_1, \dots, Y_n) > 0, \text{then } R(Y_1, \dots, Y_n) > 0$
IQR and MAD would not do the trick since they are 0 if the distribution is too discrete (but non-constant!)
Or is this impossible?
 A: This is possible in a somewhat artificial sense: just adjust $R$ a little whenever $S$ is nonzero, to guarantee $R$ is nonzero in such cases.

"Resistant" means $R$ has a finite breakdown point, but the possibility of a sample with values $(Y, Y, \ldots, Y, Y^\prime \ne Y)$ means that $R$ has to respond to the value of even the most extreme datum. It would seem that such an $R$ could not be resistant.  However, since you have imposed no other conditions on $R$ that might otherwise limit our choices, you could make it resistant by ensuring $R$ cannot change much in such cases.  For example, define it as
$$R(Y_1, \ldots, Y_n) = \text{MAD}(Y_1,\ldots, Y_n) + I(S(Y_1,\ldots,Y_n) \gt 0)/n$$
(using the indicator function $I$).  Because the amount by which this could change the MAD is bounded, $R$ has the same breakdown point as the MAD, making it (strongly) resistant.
This merely adds $1/n$ to the (non-negative) MAD whenever the SD is nonzero, guaranteeing the "magic property."  By adding a quantity that decreases to zero as the sample size increases, asymptotically this $R$ will have the same expectation as the MAD, showing that the artificial correction isn't necessarily all that bad.
Of course you would be concerned about all this only when the $Y_i$ do not have a continuous distribution or are strongly correlated (for otherwise the chance of the SD being zero would be nil). 

If you don't want to bias the estimator relative to the MAD you could, for instance, multiply the MAD by 
$$1 + \text{sign}(Y_1-Y_2)I(S(Y_1,\ldots,Y_n)\gt 0)/(2n)$$
when the $Y_i$ are iid.  (This trick obviates the need to use a randomized estimator.)  
Naturally the MAD could be replaced by almost any resistant estimator of scale.  The additive factor of $1/n$ could be replaced by any bounded nonzero function of $n$ or the multiplicative factors $1\pm 1/(2n)$ by any function with range in a finite interval $[a,b]$ and $a\gt 0$.
A: I will here gather together comments with the import that this is not possible in a simple manner. 
Consider an example such as $y = 7, 7, \cdots, 7, 42$ with two distinct values, one of which occurs only once. The singleton in this example, $42$, has outlier flavour. So in broad terms resistant measures of scale are of interest here. 
However, as @Michael M points out in the question, for such examples (with positive standard deviation) then IQR and MAD are zero. We could add that so also is the length of the shortest half. But any other alternative to the SD could hardly ignore the value of the smallest pairwise positive difference, i.e. the smallest value of $|y_i - y_j|$ that is positive. Here that is $35$ and for examples of this type it must be equal to the range, and thus quintessentially not robust (or resistant). 
Note that for binary data coded $0$ and $1$ similar situations can easily arise, but using the SD as a measure of scale is pretty much universal without discussion. For binary data the IQR and MAD would usually be 0. 
