Is there a geometric realization in integer-sided squares of $70^2 =\sum_{j=1}^{24} j^2 $? I saw this in the NAdigest mailing list,
and it was obviously suggested by
$70^2
=\sum_{j=1}^{24} j^2
$:

From: Gerhard Opfer gerhard.opfer@uni-hamburg.de
Date: November 06, 2017
Subject: Mathematics, combinatorial
Is it known, whether a square Q of size 70 x 70 can be covered
by little squares q_j of size j x j, j=1,2,...,24.
Can one say something about this problem in general.

I don't know.
My first thought
was to
look at the unit square.
However,
I realized that it was possible
to surround the unit square
with larger squares.
The fact that it is possible
to square the square
(i.e., fill an integer-sided square
with distinct integer-sided squares -
see
https://en.wikipedia.org/wiki/Squaring_the_square)
means that some property
of 70 and 1 through 24
is needed if
it is not possible.
It just might be impossible to square
a 70 x 70 square.
Your turn.
 A: According to the wikipedia link you've posted, the smallest (in terms of side length) perfect squared squares are $110\times110$, so it appears no such realization is possible.
This result is due to Ian Gambini in his doctoral thesis, and can be found here (there is no paywall!). It appears to have been determined via computer-aided search.
A: While the problem as stated is impossible, there is a solution using the third dimension.
In https://m.youtube.com/watch?v=229csazhybg a teacher proves the usual formula for the sum of consecutive squares by representing the sum as a stepped pyramid, then fitting three copies of this figure to make an almost rectangular solid.  He uses an averaging argument to get the volume of his solid and divides by three to get his sum.
The teacher misses the fact that he can combine his solid with a center-inverted version of itself to form a true rectangular block, containing six stepped pyramids and having the dimensions $(n)×(n+1)×(2n+1)$.  With this modification apply the procedure to $n=24$, thus forming a $24×25×49$ block.  Divide this block into six equal layers by cutting orthogonally to the $24$ dimension.  Each of these six layers $(4×25×49)$ has the same volume as one of the original stepped pyramids and also can be subdivided into a $2×5×7$ array of $2×5×7$ blocks.
