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Edited to add: The following is only correct if you build up your four lines from 0 (or if you only eliminate four rows at once) which is the way to get maximal points in Tetris. For a counterexample in the general case of clearing the screen in any way you like, see (and vote for) the answer by Gilad Naor

My original answer:

Color the squares black and white in a chessboard fashion and regard the parity of black squares. This is your invariant.

(The $10\times 4$ rectangle has an even number of black squares, each $\color{purple}{\text{T}}$-piece has $3$ or $1$ black squares and all other pieces clearly have $2$ black squares, so not only do you have an even number of $\color{purple}{\text{T}}$s, you have the same number of predominantly black $\color{purple}{\text{T}}$s and predominantly white $\color{purple}{\text{T}}$s.)

There is a similar further relation for the combined number of $\color{orange}{\text{L}}$s and $\color{blue}{\text{J}}$s:

The number of $\color{orange}{\text{L}}$s, $\color{blue}{\text{J}}$s and $\color{purple}{\text{T}}$s with a horizontal line of three squares is even.

Instead of the checkboard colouring, you just look at the alternate coloring of whole rows and see that these pieces are exactly the ones that contribute an odd number of black squares.

Edited to add: The following is only correct if you build up your four lines from 0 (or if you only eliminate four rows at once) which is the way to get maximal points in Tetris. For a counterexample in the general case of clearing the screen in any way you like, see (and vote for) the answer by Gilad Naor

My original answer:

Color the squares black and white in a chessboard fashion and regard the parity of black squares. This is your invariant.

(The $10\times 4$ rectangle has an even number of black squares, each $\color{purple}{\text{T}}$-piece has $3$ or $1$ black squares and all other pieces clearly have $2$ black squares, so not only do you have an even number of $\color{purple}{\text{T}}$s, you have the same number of predominantly black $\color{purple}{\text{T}}$s and predominantly white $\color{purple}{\text{T}}$s.)

There is a similar further relation for the combined number of $\color{orange}{\text{L}}$s and $\color{blue}{\text{J}}$s:

The number of $\color{orange}{\text{L}}$s, $\color{blue}{\text{J}}$s and $\color{purple}{\text{T}}$s with a horizontal line of three squares is even.

Instead of the checkboard colouring, you just look at the alternate coloring of whole rows and see that these pieces are exactly the ones that contribute an odd number of black squares.

Color the squares black and white in a chessboard fashion and regard the parity of black squares. This is your invariant.

(The $10\times 4$ rectangle has an even number of black squares, each $\color{purple}{\text{T}}$-piece has $3$ or $1$ black squares and all other pieces clearly have $2$ black squares, so not only do you have an even number of $\color{purple}{\text{T}}$s, you have the same number of predominantly black $\color{purple}{\text{T}}$s and predominantly white $\color{purple}{\text{T}}$s.)

There is a similar further relation for the combined number of $\color{orange}{\text{L}}$s and $\color{blue}{\text{J}}$s:

The number of $\color{orange}{\text{L}}$s, $\color{blue}{\text{J}}$s and $\color{purple}{\text{T}}$s with a horizontal line of three squares is even.

Instead of the checkboard colouring, you just look at the alternate coloring of whole rows and see that these pieces are exactly the ones that contribute an odd number of black squares.

Edited to add: The following is only correct if you build up your four lines from 0 (or if you only eliminate four rows at once) which is the way to get maximal points in Tetris. For a counterexample in the general case of clearing the screen in any way you like, see (and vote for) the answer by Gilad Naor

My original answer:

Color the squares black and white in a chessboard fashion and regard the parity of black squares. This is your invariant.

(The $10\times 4$ rectangle has an even number of black squares, each $\color{purple}{\text{T}}$-piece has $3$ or $1$ black squares and all other pieces clearly have $2$ black squares, so not only do you have an even number of $\color{purple}{\text{T}}$s, you have the same number of predominantly black $\color{purple}{\text{T}}$s and predominantly white $\color{purple}{\text{T}}$s.)

There is a similar further relation for the combined number of $\color{orange}{\text{L}}$s and $\color{blue}{\text{J}}$s:

The number of $\color{orange}{\text{L}}$s, $\color{blue}{\text{J}}$s and $\color{purple}{\text{T}}$s with a horizontal line of three squares is even.

Instead of the checkboard colouring, you just look at the alternate coloring of whole rows and see that these pieces are exactly the ones that contribute an odd number of black squares.

Color the squares black and white in a chessboard fashion and regard the parity of black squares. This is your invariant.

(The $10\times 4$ rectangle has an even number of black squares, each T-piece has 3 or 1 black squares and all other pieces clearly have 2 black squares, so not only do you have an even number of Ts, you have the same number of predominantly black Ts and predominantly white Ts.)

There is a similar further relation for the combined number of Ls and Js:

The number of Ls, Js and Ts with a horizontal line of three squares is even.

Instead of the checkboard colouring, you just look at the alternate coloring of whole rows and see that these pieces are exactly the ones that contribute an odd number of black squares.

Color the squares black and white in a chessboard fashion and regard the parity of black squares. This is your invariant.

(The $10\times 4$ rectangle has an even number of black squares, each $\color{purple}{\text{T}}$-piece has $3$ or $1$ black squares and all other pieces clearly have $2$ black squares, so not only do you have an even number of $\color{purple}{\text{T}}$s, you have the same number of predominantly black $\color{purple}{\text{T}}$s and predominantly white $\color{purple}{\text{T}}$s.)

There is a similar further relation for the combined number of $\color{orange}{\text{L}}$s and $\color{blue}{\text{J}}$s:

The number of $\color{orange}{\text{L}}$s, $\color{blue}{\text{J}}$s and $\color{purple}{\text{T}}$s with a horizontal line of three squares is even.

Instead of the checkboard colouring, you just look at the alternate coloring of whole rows and see that these pieces are exactly the ones that contribute an odd number of black squares.