Define, for all positive integer $n$ :
$$S_n=\sum_{1\leqslant i,j\leqslant n}\min(i,j)$$
A simple calculation leads to the following closed formula :
$$\forall n\in\mathbb{N}^\star,\,S_n=\frac{n(n+1)(2n+1)}6$$which is notoriously equal to $\displaystyle{T_n=\sum_{k=1}^nk^2}$.
My question : Is it possible to anticipate the relation $\forall n\in\mathbb{N}^\star,\,S_n=T_n$ without doing any computation ?
Yes, this is expected.
Let's stack some cubes of $n$ layers with two sides at a corner of a room.
At the bottom layer, we have a square of size $n \times n$.
At the second bottom layer, we have a square of size $(n-1) \times (n-1)$.
For the case of $k$-th level from the top, we have a square of size $k \times k$.
Source of picture
For small cases:
We have
$$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3\end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{bmatrix}+ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{bmatrix}+ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
The matrix of the left contains $\min(i,j)$ where the terms on the right show the layers of cubes.
It is counting by columns vs counting by layers.
Consider the following $n\times n$ grid
$$\begin{array}{ccc} 1 &1 &1 &1 &\ldots &1\\ 1 &2 &2 &2 &\ldots &2\\ 1 &2 &3 &3 &\ldots &3\\ 1 &2 &3 &4 &\ldots &4\\ \vdots &\vdots &\vdots &\vdots &\ddots &\vdots\\ 1 &2 &3 &4 &\ldots &n \end{array}$$
where the $(i,j)$-th entry is $\min\{i,j\}$.
Now strip off each layer (of $1$s and then $2$s and so on) and sum them separately. That will give $$1\times (2n-1) + 2\times (2n-3) + 3\times (2n-5) + \dots + n\times 1$$ which can be re-interpreted as $$(2n-1) + (2n-3) + (2n-3) + (2n-5) + (2n-5) + (2n-5) + \dots$$ and since the sum of the first $n$ odd integers is $n^2$, you have your proof.
This answer generalizes the question.
Call the sum $S$ for the case asked. Consider a tuple $(a,b) \in \mathbb N^2$ where $1\le a,b\le n$. Consider some $r \in [1,n] \cap \mathbb N$. For how many tuples $(a,b)$ is $\min(a,b) = r$? You can construct exactly $(n-r+1)^2 - (n-r)^2$ such tuples$^{*}$. Thus the sum is just
$$\begin{align} S &= \sum_{r=1}^{n} r((n-r+1)^2 - (n-r)^2) \\ &= \sum_{r=1}^{n} r(n-r+1)^2 - \sum_{r=1}^{n} r(n-r)^2 \\ &= \sum_{r=1}^{n} r(n-r+1)^2 - \sum_{\color{red}{r=0}}^{n} r(n-r)^2\\ &= \sum_{r=1}^n (n-r+1)r^2 - \sum_{r=0}^n (n-r)r^2\tag{1.1}\\ &= \sum_{r=1}^n (n-r+1)r^2 - \sum_{\color{red}{r=1}}^n (n-r)r^2\\ &= \sum_{r=1}^n r^2 \end{align}$$
REMARKS:
$^{*}$: the most important aspect of the solution is counting the number of tuples where $\min(a,b)=r$ is obeyed. To count them first count all the tuples where $(a,b) \in \{r,r+1,r+2,\dots, n\}^2$ and then subtract all tuples where $(a,b) \in \{r+1,r+2,r+2, \dots, n\}^2$. This counts all tuples where $\min(a,b) \geq r$ and then subtracts all tuples where $\min(a,b)\geq r+1$.
$(1.1)$ uses the fact that $\sum_{r=a}^{n} f(r) = \sum_{r=a}^n f(n-r+a)$. This is called reversing the order of the sum.
I provide two bonuses;
Claim $1$: $S_1(k,n) = \underset{1\le x_1,x_2,x_3,\dots,x_k \le n}{\sum} \min(x_1,x_2,x_3,\dots, x_k) =\sum_{r=1}^n r^k$
Setting $k=2$ in this case yields the desired identity asked in the question.
Proof: Consider all $k$-tuples $\in \mathbb N^k$ where all their elements are in $[1,n]$. There are exactly $(n-r+1)^k - (n-k)^k$ such $k$-tuples which have their minimum element some fixed $r$. Since $r\in [1,n] \cap \mathbb N$ we have
$$\begin{align} S_1(k,n) &= \sum_{r=1}^n r((n-r+1)^k - (n-r)^k) \\ &= \sum_{r=1}^{n} (n-r+1)(r)^k - \sum_{r=1}^n (n-r)(r)^k \\ &= \boxed{\sum_{r=1}^n r^k} \end{align}$$
Claim $2$: $S_2(k,n) = \underset{1\le x_1,x_2,x_3,\dots,x_k \le n}{\sum} \max(x_1,x_2,x_3,\dots, x_k) =n^{k+1} - \sum_{r=1}^{n-1} r^k$
Proof: As usual, there are exactly $r^k - (r-1)^k$ $k$-tuples such that the tuple's maximal element is some $r\in[1,n]\cap \mathbb N$ where the $k$-tuples are $\in \mathbb N^k$ where each element $\in [1,n]$. Thus $$\begin{align} S_2(k,n) &= \sum_{r=1}^n r(r^k - (r-1)^k)\\ &= n^{k+1}+\sum_{r=1}^{n-1}r^{k+1} - \sum_{r=1}^{n-1}(r+1)r^k\\ &= \boxed{n^{k+1}-\sum_{r=1}^{n-1} r^k} \end{align}$$
For a fixed $k$, there are precisely $2(n-k+1) - 1$ tuples $(i,j) \in [n] \times [n]$ such that $\min(i,j) = k$. Thus, the sum can be expressed as
$$\sum_{k=1}^n k \times (2(n-k+1) - 1) = \sum_{k=1}^n (n-k+1) \times (2k - 1) $$
A number $2k-1$ is contributing to this sum $(n-k+1)$ times. We can visualize this as a stacked sequence of rows of decreasing length where row $1$ has $n$ repetitions of $1$, row $2$ has $n-1$ repetitions of $3$ and so on.
The result $S_n = T_n$ is apparent by instead summing this construction column-wise and recalling that the sum of the first $k$ odd numbers is $k^2$. The latter fact can be shown purely combinatorially.
