Rank of smoothing matrix The solution to the minimizing problem: $$RSS(f, \lambda) = \sum_{i=1}^n (y_i - f(x_i))^2 + \lambda \int (f''(t))^2 dt$$ is written in matrix form as $f = N\theta = \sum_{j = 1}^n N_j(x_i) \theta_j$ where the $N_j(x)$ are a $n$-dimensional set of basis functions representing a natural cubic spline. Together with a choice of $\lambda$ we have a smoothing spline with the estimate $$\hat f = S_\lambda y$$ where $$S_\lambda = N(N^tN + \lambda \Sigma_N)^{-1}N^t$$ and $\Sigma_{N_{ij}} = (\int N''_i(t)N''_j(t) dt)_{ij}$ and $N_{ij} = N_j(x_i)$.

Why is $\mathrm{Rank}(S_\lambda) = n$, or equivalently $N$ invertible?

This is a partial answer to be completed, I posted it just for clarification and drawing more attention.
I have consulted the ESL section.  Let me first clarify the notations.  The problem is 1D function estimation so the training data is $(x_i, y_i): i = 1, \ldots, n$ (as I pointed out in the comment, using $N$ to denote the number of observations is an unfortunate confusion, even in ESL the spline matrix is boldfaced.  So here I will still use $n$ to denote the number of observations).  The spline matrix $N$ is defined as
\begin{align}
N = \begin{bmatrix}
N_1(x_1) & N_2(x_1) & \cdots & N_n(x_1) \\
N_1(x_2) & N_2(x_2) & \cdots & N_n(x_2) \\
\vdots   & \vdots   & \ddots & \vdots   \\
N_1(x_n) & N_2(x_n) & \cdots & N_n(x_n)
\end{bmatrix}. \tag{1}
\end{align}
Evidently, if $x_i = x_j$ for some $i \neq j$, then clearly $N$ is not invertible (because then the $i$-th row and the $j$-th row are identical).  So in order $\operatorname{rank}(S_\lambda) = n$ (a claim made by ESL, p.153), $x_1, \ldots, x_n$ must be distinct (without loss of generality, assume $x_1 < x_2 < \cdots < x_n$).
According to the author:

$N_j(x)$ are an $n$-dimensional set of basis functions for representing this family of natural splines...

By calling $\{N_1(x), \ldots, N_n(x)\}$ "basis functions" (to be fair, doing so without first proving they really formed a basis of a vector space -- which was not clearly stated in advance, is loose), they are necessarily linearly independent in the vector space $V$ of spline functions (known as "spline space", more accurately, given knots $x_1 < x_2 < \cdots < x_n$ and only cubic splines are under consideration, I think $V$ is a real vector space spanned by
$\{1, x, x^2, x^3, (x - x_1)_+^3, \ldots, (x - x_n)_+^3\}$).  As you are interested in, this statement, of course requires a mathematical proof, which the text omitted.
Recall $N_j(x)$ is defined as (Equation (5.4) in ESL):
\begin{align}
N_1(x) = 1, N_2(x) = x, N_{k + 2}(x) = d_k(x) - d_{n - 1}(x), k = 1, \ldots, n - 2,
\tag{2}
\end{align}
where
\begin{align}
d_k(x) = \frac{(x - x_k)_+^3 - (x - x_n)_+^3}{x_n - x_k}. \tag{3}
\end{align}
Plug $(2)$ and $(3)$ into $(1)$, then $\det(N)$ (denoted by $D_n$) is
\begin{align}
& D_n = \begin{vmatrix}
1 & x_1 & 0                                   & 0 &\cdots & 0 \\      
1 & x_2 & (x_2 - x_1)^3/(x_n - x_1)           & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots                      & \vdots & \ddots & \vdots \\
1 & x_{n - 1} & (x_{n - 1} - x_1)^3/(x_n - x_1) & (x_{n - 1} - x_2)^3/(x_n - x_2) & 
\cdots & (x_{n - 1} - x_{n - 2})^3/(x_n - x_{n - 2}) \\
1 & x_n & (x_n - x_1)^2 - (x_n - x_{n - 1})^2 & (x_n - x_2)^2 - (x_n - x_{n - 1})^2 & \cdots & (x_n - x_{n - 2})^2 - (x_n - x_{n - 1})^2
\end{vmatrix} \\
& ???? \neq 0.
\end{align}
When $n = 1, 2, 3$,
\begin{align}
& D_1 = 1 \neq 0, \\
& D_2 = x_2 - x_1 \neq 0, \\
& D_3 = 2(x_2 - x_1)^2(x_3 - x_2) \neq 0,
\end{align}
all confirming $N$ is invertible.  However, for $n \geq 4$, it is more difficult to make a conclusion.  I have calculated
\begin{align}
D_4 &= (x_4 - x_1)^{-1}(x_2 - x_1)(x_3 - x_2)^2(x_2 + x_3 - 2x_1)(2x_4 - x_2 - x_3) \\
&+ (x_4 - x_2)^{-1}(x_2 - x_1)(x_3 - x_2)^3[(x_1 - x_2)^2 + (x_3 - x_4)^2 - (x_4 - x_1)^2],
\end{align}
but it is not immediate to see if $D_4 \neq 0$.
