What are the consequences of "copying" a data set for OLS? Suppose I have a random sample $\lbrace X_i, Y_i\rbrace_{i=1}^n$. Assume this sample is such that the Gauss-Markov assumptions are satisfied such that I can construct an OLS estimator where
$$\hat{\beta}_1^{OLS} = \frac{\text{Cov}(X,Y)}{\text{Var(X)}}$$
$$\hat{\beta}_0^{OLS} = \bar{Y} - \bar{X} \hat{\beta}_1^{OLS}$$
Now suppose I take my data set and double it, meaning there is an exact copy for each of the $n$ $(X_i,Y_i)$ pairs.
My Question
How does this affect my ability to use OLS? Is it still consistent and identified? 
 A: Do you have a good reason to do the doubling (or duplication?) It doesn't make much statistical sense, but still it is interesting to see what happens algebraically.  In matrix form your linear model is 
$$ \DeclareMathOperator{\V}{\mathbb{V}}   
   Y = X \beta + E,
$$
the least square estimator is $\hat{\beta}_{\text{ols}} = (X^T X)^{-1} X^T Y $  and the variance matrix is $ \V \hat{\beta}_{\text{ols}}= \sigma^2 (X^t X)^{-1} $.  "Doubling the data" means that $Y$ is replaced by $\begin{pmatrix} Y \\ Y  \end{pmatrix}$ and $X$ is replaced by  $\begin{pmatrix} X \\ X  \end{pmatrix}$. The ordinary least squares estimator then becomes 
$$
   \left(\begin{pmatrix}X \\ X \end{pmatrix}^T \begin{pmatrix} X \\ X \end{pmatrix} \right )^{-1} \begin{pmatrix} X \\ X  \end{pmatrix}^T \begin{pmatrix} Y \\ Y  \end{pmatrix}  =   \\
(x^T X + X^T X)^{-1} (X^T Y + X^T Y ) = (2 X^T X)^{-1} 2 X^T Y = \\
\hat{\beta}_{\text{ols}}
$$
so the calculated estimator doesn't change at all. But the calculated variance matrix becomes wrong: Using the same kind of algebra as above, we get the variance matrix $\frac{\sigma^2}{2}(X^T X)^{-1}$, half of the correct value.  A consequence is that confidence intervals will shrink with a factor of $\frac{1}{\sqrt{2}}$.
The reason is that we have calculated as if we still have iid data, which is untrue: the pair of doubled values obviously have a correlation equal to $1.0$.  If we take this into account and use weighted least squares correctly,  we will find the correct variance matrix. 
From this, more consequences of the doubling will be easy to find as an exercise, for instance, the value of R-squared will not change. 
A: I am not yet familiar enough with the theory to give you a very mathematical answer, but intuitively, OLS only cares about proportions in which different cases are present. This makes sense when you recall that OLS chooses the coefficients that minimize the mean of the squared residuals, and the mean reflects purely the proportions of its inputs (in the sense that the mean of (1, 3, 3) is the same as the mean of a dataset with a million 1s and two million 3s). So, doubling the dataset will get you the identical model.

Here's an R example, where we generate a random regression problem and notice that the coefficients are unchanged when doubling the data:
nc = sample(1:10, 1, replace = T)
n = sample(11:500, 1, replace = T)
x = as.matrix(replicate(nc, rnorm(n)))
coef = rnorm(nc)
sd.resid = runif(1, 0, 5)

y = x %*% matrix(coef) + rnorm(n, sd = sd.resid)

print(cbind(
  coef(lm(y ~ x)),
  coef(lm(c(y, y) ~ rbind(x, x)))))

One run gives me:
                   [,1]        [,2]
(Intercept) -0.10002238 -0.10002238
x1          -2.14801619 -2.14801619
x2           0.23120764  0.23120764
x3           0.05360792  0.05360792
x4           1.91972198  1.91972198
x5          -1.09887264 -1.09887264
x6           0.04248358  0.04248358

