Is IPTW (inverse probability of treatment weighting) legal? When using IPTW, one can easily get weights 10 or even 20 for the observations.
For instance, in logistic regression, weight 10 for an observation means that we have not one, but 10 observations identical to this one. Thus, if we are allowed to assign weights > 1 to observations, then we can take, e.g., a sample of 100 observations, assign weight 10 to each one of them, and thus obtain a sample of 1000 observations, with totally lesser p-values etc.. But that will be evidently illegal.
Why assigning weights > 1 is legal in IPTW? Is IPTW with weights >1 legal?
 A: That is not how it works. The inference based on logistic regression is not correct when you incorporate weighting. You need to estimate the variance of the IPTW estimator, which happens to be inversely related to the propensity score. So large weights also lead to large variance estimates and thus larger p-values.
(Also, with IPTW, all weights are larger than one since it is the inverse of a probability).
Here is a ultra mini-lesson on IPW estimators. Suppose you observe the data structure $(X,A,Y)$ where $X$ is a vector of covariates, $A$ is a binary treatment, and $Y$ is some outcome. Let $\pi_0(x):= P(A=1|X=x)$ be the propensity score. Suppose we are interested in estimating the treatment-specific mean parameter $Psi := E_XE[Y|A=1,X]$. Consider the identity
$$E_XE[Y|A=1,X] = E_X \left[\frac{E[Y|A,X]1(A=1)}{\pi_0(X)}\right] =  E_X \left[\frac{Y1(A=1)}{\pi_0(X)}\right],$$
which follows from a conditioning argument. This suggests the following IPW estimator of $\Psi$:
$$\hat \Psi_n := \frac{1}{n} \sum_{i=1}^n \frac{Y_i1(A_i=1)}{\pi_0(X_i)}$$
where we unrealistically assume that $\pi_0$ is known. Since $\hat \Psi_n$ is just an average of a random variable, inference is easy. We have
$$\sqrt{n}(\hat \Psi_n - \Psi) \rightarrow_d  N\left(0, \text{Var}_0\{Y1(A=1)/\pi_0(X)\} \right).$$
Now, note that the variance depends on the weight $w_0=\frac{1}{\pi_0(X)}$, so a larger weight gives a larger variance.
You are free to replace $\pi_0$ with a logistic regression estimate of $\pi_n$, but this is usually not a $\sqrt{n}$-consistent or asymptotically normal estimator without very strong conditions.
Note, $E[Y|A=1]$ and $E_XE[Y|A=1,X]$ are two very different parameters.
