As a side note, the excessive memory usage in the original code can be avoided by using instead

RRmodel = 1. - Exp[-(x/dm)^n];

that is, replacing the exact number 1 with a machine precision 1.

Whenever a tiny number is subtracted from an exact 1, the result will have too many digits of precision, consider for example

1 - Exp[-100000.]

Adding a few more zeros to the Exp argument can make any computer run out of memory.

This behavior may also be circumvented by telling the kernel to no longer trap machine underflow and thus avoid switching to high-precision arithmetic:

In[1]:= SetSystemOptions["CatchMachineUnderflow" -> False]; 

In[2]:= 1 - Exp[-100000.]  

Out[2]= 1.

In fact, the above may become the default (and only) behavior for machine arithmetic in a future release.

If the raw data isn't available, then one can still obtain maximum likelihood estimates from the binned data (assuming the binned data can be transformed into the observed frequency counts).

sizedistcum = {{1.12996, 0.}, {1.42366, 0.}, {1.7937, 0.}, {2.25992, 
    0.}, {2.84732, 0.}, {3.5874, 0.}, {4.51984, 0.}, {5.69464, 
    0.}, {7.1748, 0.}, {9.03968, 0.}, {11.3893, 0.}, {14.3496, 
    0.}, {18.0794, 0.}, {22.7786, 0.}, {28.6992, 0.}, {36.1587, 
    0.}, {45.5572, 0.}, {57.3984, 0.00215924}, {72.3175, 
    0.0283401}, {91.1143, 0.0537112}, {114.797, 0.0753036}, {144.635, 
    0.103104}, {182.229, 0.268826}, {229.594, 0.832659}, {289.27, 
    0.964372}, {364.457, 0.993522}, {459.187, 1.}, {578.54, 
    1.}, {728.914, 1.}, {918.375, 1.}, {1157.08, 1.}, {1457.83, 
    1.}, {1836.75, 1.}, {2314.16, 1.}, {2915.66, 1.}, {3673.5, 
    1.}, {4628.32, 1.}, {5831.32, 1.}, {7347., 1.}, {9256.64, 
    1.}, {11662.6, 1.}};

(* Frequency counts for each bin *)
c = Differences[Round[3705 sizedistcum[[All, 2]]]]
(* Bin borders *)
x = N[PowerRange[1, 13000, 2^(1/3)]]

(* Cumulative distribution function *)
F[x_] := 1 - Exp[-(x/dm)^n]

(* Log of likelihood *)
logL = Sum[c[[i]] Log[F[x[[i + 1]]] - F[x[[i]]]], {i, Length[c]}];

$$ \log {L} = \sum_ {i = 1}^k c_i \log {(F (x_ {i + 1}) - F (x_i))}$$

(* Find maximum likelihood estimates *)
sol = FindMaximum[logL, {{n, 4}, {dm, 200}}, WorkingPrecision -> 20]
(* {-5891.8004314673819211, {n -> 4.3748213953340505200, dm -> 191.19015070319332793}} *)

(* Show results *)
Show[ContourPlot[logL, {n, 3, 6}, {dm, 150, 250}, PlotRange -> All,
  Contours -> -{6, 7, 8, 9, 10, 11, 12} 1000, 
  PlotLabel -> Style["Log likelihood surface", Black, 24],
  Frame -> True, 
  FrameLabel -> (Style[#, Bold, Italic, 18] &) /@ {"n", "dm"}],
 ListPlot[{{n, dm}} /. sol[[2]], PlotStyle -> Red]]

(* Standard errors *)
cov = -Inverse[(D[logL, {{n, dm}, 2}]) /. sol[[2]]];
sen = cov[[1, 1]]^0.5
(* 0.05702 *)
sedm = cov[[2, 2]]^0.5
(* 0.786466 *)

To repeat: this only works if one can reconstruct the counts as the total count is the sample size rather than the number of bins (which is completely artificial).

Thank you for the detailed description about the aplication of "WeibullDistribution" and raw data. The Rosin-Rammler model is used to describe Particle-size distributions. In fact, the raw data containing the size of individual particles could be obtained from certain experimental measurements, eg: image analysis.
In other hand, particle size distributions that were obtained from mesh sieving and laser difraction will show the size distribution described as "histogram binning". In this last case, the model fitting is the most used tool.

I would like to hear another opinion for this statistical atrocity. I'll remember that my question is about the "memory error" and freezing of Mathematica kernel.

I'm sorry I wasn't convincing in my attempt to keep you from completing a statistical atrocity. This time I'll suggest how the appropriate approach is much simpler:

First one should use the raw data and not do any histogram binning. Here a random sample is chosen from a Rosin-Rammler (Weibull) distribution and the FindDistributionParameters function is used to obtain the maximum likelihood parameter estimates:

SeedRandom[12345];
x = RandomVariate[WeibullDistribution[4.5, 200], 3705];
mle = FindDistributionParameters[x, WeibullDistribution[n, dm]]
(* {n -> 4.57036, dm -> 199.591} *)

That's all there is to it. Well, other than constructing a measure of precision for the estimates (which is essential). That only takes a few more steps. Below are the steps to obtain the standard errors for the parameter estimates:

logL = LogLikelihood[WeibullDistribution[n, dm], x];
cov = -Inverse[(D[logL, {{n, dm}, 2}]) /. mle];
sen = cov[[1, 1]]^0.5
(* 0.0584913 *)
sedm = cov[[2, 2]]^0.5
(* 0.755647 *)

Performing a regression is bad news because of the many assumptions that are violated: (1) This is not a regression situation but rather a situation fitting a probability distribution, (2) the assumption of constant variance that NonlinearModelFit assumes is violated, (3) In this case one has 3705 observations and not the artificial sample size of an arbitrary binning. And there are certainly other violations. (I will admit that maybe a regression approach could be used to get starting values for the maximum likelihood estimates. But that's all it should be used for.)

Given that the dataset name (sizedistcum) has "dist" and "cum" in it and it appears that you have equal intervals with respect to the log of x, I suspect you have a sample cumulative distribution function formed from a binning of a random sample of size 3,705. The function that you're trying to fit is a cumulative distribution function for a Weibull distribution. If so, performing a regression is inappropriate.

The counts for the histogram seem to be

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 97, 94, 80, 103, 614, 2089, 488, 108, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

If so, I can show you how to estimate the parameters of the Weibull distribution from a histogram if my speculation about your data is correct. But first I want to determine if my speculation is correct.

Jim