Proof for "The sum of the observed values $Y_i$ equals the sum of the estimated / fitted values $\hat Y_i$" I needed some help trying to understand why the sum of the observed values $Y_i$ equals the sum of the estimated values $\hat{Y}_i$. 
 A: $Y_i = \hat{Y}_i + \hat{\epsilon_i}$ by definition.
Also, we know that $\frac{1}{n}\sum_{i=1}^{n}{\hat{\epsilon}_i}=0$ because the intercept of the model absorbs the mean of the residuals.
So, $\frac{1}{n}\sum_{i=1}^{n}{Y_i} = \frac{1}{n}\sum_{i=1}^{n}{\hat{Y}_i} + \frac{1}{n}\sum_{i=1}^{n}{\hat{\epsilon_i}} = \frac{1}{n}\sum_{i=1}^{n}{\hat{Y}_i} + 0 = \frac{1}{n}\sum_{i=1}^{n}{\hat{Y}_i}$
A: Let $P$ be the projection matrix on $X$, where one of the columns of $X$ is $\mathbf{1}$, where $\mathbf{1}$ is vector of ones, then
\begin{align}
\mathbf{1}'\hat{y}&=\mathbf{1}'Py\\
&=\left(P'\mathbf{1}\right)'y\\
&=\mathbf{1}'y.
\end{align}
A: Let us consider the special case of a simple linear regression.
Minimizing the sum of squared residuals $S=\sum_{r=1}^{n}(y_{r}-a-b x_{r})^{2}$ w.r.t. $a$ and $b$ gives the "normal equations" (first order conditions)
$$
\sum_{r=1}^{n}y_{r}=na+b\sum_{r=1}^{n}x_{r}\quad\text{and}\quad\sum_{r=1}^{n}y_{r}x_{r}=a\sum_{r=1}^{n}x_{r}+b\sum_{r=1}^{n}x_{r}^{2}.
$$
The fitted values are $\hat{y}_{r}=a+bx_{r}$, with residuals $$\hat{u}_{r}=y_{r}-(a+bx_{r})$$
Consider
$$\sum_{r=1}^{n}\hat{u}_{r}=\sum_{r=1}^{n}y_{r}-\sum_{r=1}^{n}\hat{y}_{r}=\sum_{r=1}^{n}y_{r}-\sum_{r=1}^{n}(a+bx_{r})=\sum_{r=1}^{n}y_{r}-na-b\sum_{r=1}^{n}x_{r}.$$
Now, substitute $\sum_{r=1}^{n}y_{r}$ for the first normal equation.
