Can a 10*10 square be paved with 1*4 rectangular stone plates?

A very simple approach is to tile the 10x10 square

with alternating 2x2 squares of +1 and -1:

In the picture +1 and -1 are abbreviated + and -:

    + + - - + + - - + +
    + + - - + + - - + +
    - - + + - - + + - -
    - - + + - - + + - -
    + + - - + + - - + +
    + + - - + + - - + +
    - - + + - - + + - -
    - - + + - - + + - -
    + + - - + + - - + +
    + + - - + + - - + +
 

Every 1x4 or 4x1 tile covers

2 + and 2 -. The net total under each such tile is therefore zero and the same must hold for the area covered by any number of 1x4 and 4x1 tiles.

But there are 5x5 = 25, an odd number, of 2x2 squares meaning that they can't split evenly between + and -. The overall sum cannot be zero (it is +4) and hence a tiling cannot exist.

The following feels intuitive enough to me. Colour the grid as follows.

 R G B Y R G B Y R G
 Y R G B Y …
 B Y R G B …
 …

It’s clear that whichever way you place a 1 × 4 tile, it covers all four colours. But the number of R cells is larger than the number of B cells, so the grid can’t be tiled.

A proof without a colouring is as follows:

Every row must intersect 2,6 or 10 vertical tiles. In particular, the number of vertical tiles is even.

By the same token the number of horizontal tiles is even. But the total number of tiles must be 25, an odd number and a contradiction.

Addendum: @Mark Amery kindly expands on the first step noting it "[...] left a lot of work to the reader [...]".

[B]ase case[:] [...] the number of vertical tiles starting on the bottommost "rung" (i.e. starting from the bottom of the grid) must be even. [A]nd then we apply [...] as an inductive step [...] that the number of vertical tiles starting from every other rung is also even [...] since it has an even difference from the even number on the previous [but three] rung. [W]e conclude that the total number of vertical tiles, as a sum of even numbers, must be even.

Use the Integer-Rectangle Tiling Lemma:

Whenever a rectangle is tiled by rectangles each of which has at least one integer side, then the tiled rectangle has at least one integer side.

Scale down the sides in the puzzle from "Can a 10×10 square be tiled with 1×4 rectangles?" to "Can a 2.5×2.5 square be tiled with 0.25×1 rectangles?

Answer: NO (the big rectangle has no integer sides, but the small rectangles all do)

Here's an approach via linear programming, where the dual variables suggest a coloring that automatically yields a certificate of infeasibility.

For each possible $1 \times 4$ rectangle $r$, let binary decision variable $x_r$ indicate whether that rectangle is used. For each cell $(i,j)$, let binary decision variable $y_{ij}$ indicate whether that cell is uncovered. Let $R_{ij}$ be the set of rectangles that cover cell $(i,j)$. The problem is to minimize $$\sum_{i=1}^{10} \sum_{j=1}^{10} y_{ij}$$ subject to linear constraints \begin{align} \sum_{r \in R_{ij}} x_r + y_{ij} &= 1 &&\text{for $(i,j)\in [10]\times[10]$} \end{align} The minimum turns out to be $4$, meaning that $4$ cells are uncovered. The resulting optimal dual variables are:

In this coloring, every $1 \times 4$ rectangle has sum $0$, but the entire grid has sum $4$ and so cannot be partitioned. Or just forget the numbers and note that each rectangle contains exactly one blue square; we need $25$ rectangles, but there are only $24$ blue squares.

An alternative optimal dual solution is:

As in the previous coloring, for this coloring (which is the same as the one in the answer by @Albert.Lang), every $1 \times 4$ rectangle has sum $0$, but the entire grid has sum $4$ and so cannot be partitioned. Or just forget the numbers and note that each rectangle contains exactly two blue squares; we need $25$ rectangles ($50$ blue squares), but there are only $48$ blue squares.

Pranay answers the question, but upon further review you don't need an exact count of each color square. You need only to distinguish even from odd, making this a "parity check" puzzle.

Using Pranay's pattern

 R G B Y R G B Y R G
 Y R G B Y …
 B Y R G B …
 …

we note that

the diagonal rows of R squares each contain an even number (the longest such row is ten and others come in decrements of four), but to cover the grid we would need an odd number (25) of tiles each covering one R square.

The following well-known solution is so simple that it even does not need a coloring. :-)

Let $A$ be the square $[0,10]\times [0,10]$. Suppose for a contradiction that $A$ can be paved with $1\times 4$ rectangular plates $A_1,\dots, A_n$. Put $f(x,y)=\sin \frac\pi{2} x\cdot\sin \frac\pi{2} y$ for each $x,y\in A$.

Then $$\int\int_A f(x,y)\, dx\, dy=\int_0^{10}\sin \frac\pi{2} x\, dx\cdot \int_0^{10} \sin \frac\pi{2} y\, dy =$$ $$\left(\frac {(-2)}{\pi}\cdot \cos\frac\pi{2} t{\Big|}_0^{10}\right)^2=\left(\frac {(-2)}{\pi}(\cos 5\pi-\cos 0)\right)^2=\left(\frac 4{\pi}\right)^2\ne 0.$$

On the other hand, we can similarly show that for each natural $i\le n$ we have $$\int\int_{A_i} f(x,y)\, dx\, dy=0.$$

But then $$0\ne \int\int_A f(x,y)\, dx\, dy = \int\int_{\bigcup_{i=1}^n A_i} f(x,y)\, dx\, dy=$$ $$\sum_{i=1}^n \int\int_{A_i} f(x,y)\, dx\, dy = \sum_{i=1}^n 0=0,$$ a contradiction.

Simpler coloring, but slightly more complicated reasoning:

Color as follows:

 R G B Y R G B Y R G
 R G B Y R G B Y R G
 R G B Y R G B Y R G
 …

Reasoning:

There are 30 each of Red and Green, and 20 each of Blue and Yellow. Now we reduce to a single player game of Nim, where legal moves are to reduce each color by 1, or reduce one color by 4. The goal is to reduce all colors to 0.

The only way to change the count of one color modulo 4 is to change all of them at once by the same amount. 20 is divisible by 4, whereas 30 is not. Therefore, the tiling is impossible.

Here's a meta-reasoning approach that uses the postscript to find extremely intuitive solution. It's noted that the original source provides a solution which is

not intuitive. But if the tiling could be accomplished, the answer would simply be a picture of it. Knowing that the solution of the problem is "non-intuitive" means that we cannot simply draw a picture of it, so therefore the solution is not a configuration where the tiling is complete, but rather a proof that it can't be done.