For every x, y ∈ IN = {1, . . . , N} define K(x, y) = min{x, y}. Prove that K is a valid kernel I need to prove K is a valid Kernel.
x and y are positive integers, and K is the minimum of x and y.
I know I should prove K positive definite. But May I see how.

Let $\mathcal X$ be any subset of $\{1...N\}$, and $|\mathcal X|=n,$ and $x_i,x_j$ be any two elements from this set.
We can write $\min(x_i,x_j)$ as below:
$$\min(x_i,x_j)=\int_0^N \mathbb 1(t\leq x_i)\mathbb 1(t\leq x_j)dt$$
The reason of this equality is straightforward. The first indicator function will be $1$ until $x_i$; and the second indicator function is $1$ until $x_j$. The multiplication will be $1$ if both are $1$, which means until the minimum of $x_i$ and $x_j$. The integral then can be written as
$$\begin{align}\int_0^N\mathbb 1(t\leq x_i)\mathbb 1(t\leq x_j)dt&=\int_0^{\min(x_i,x_j)}\mathbb 1(t\leq x_i)\mathbb 1(t\leq x_j)dt\\&=\int_0^{\min(x_i,x_j)}1\times 1\times dt\\&=\min(x_i,x_j)\end{align}$$
Then, we substitute in to the PSD requirement:
$$\begin{align}\sum_{i=1}^n\sum_{j=1}^n c_i c_j \min(x_i,x_j)&=\sum_{i=1}^n\sum_{j=1}^n c_i c_j \int_0^N \mathbb 1(t\leq x_i)\mathbb 1(t\leq x_j)dt \\&=\int_0^N \underbrace{\left(\sum_{i=1}^n c_i\mathbb 1(t\leq x_i)\right)}_{M(t)}\underbrace{\left(\sum_{j=1}^n c_j\mathbb 1(t\leq x_j)\right)}_{M(t)}dt\\&=\int_0^NM(t)M(t)dt=\int_0^NM(t)^2dt\geq 0\end{align}$$
Both multiplicands are equal because both summations range from $1$ to $n$. We can also change the order of summations and integrations according to Fubini's theorem (in short, the sums and integrals are finite).
This covers PSD property and you've already mentioned the symmetric property in your comment.
