Imagine the numbers from $1$ to $10$ written in clockwise order around a circle. I will call the clockwise distance from one number to another the "step" of that pair of numbers. For example, for the pair $(2,5)$ the step is $3$, just the ordinary difference $5-2$. But for $(8,7)$, the step is $9$, because we're going the long way around.
In order for the tenth pair in a box to end with the same number as the first pair started with, the total of the steps for all ten pairs must be a multiple of $10$. There are ten pairs with each possible step size from $0$ to $9$, so one might try putting all ten pairs with a given step in the same box. For example, the ten pairs with step size $3$ could be arranged in a box like this:
$$
\big[\ (1,4)\quad (4,7)\quad (7,10)\quad (10,3)\quad (3,6)\quad
(6,9)\quad (9,2)\quad (2,5)\quad (5,8)\quad (8,1)\ \big]
$$
This works fine for some steps (viz., $1$, $3$, $7$ or $9$), but not for others. For example, if we gather the pairs with step $5$ together and put $(1,6)$ in the box first, the second pair would be $(6,1)$, but the third would have to be $(1,6)$ again.
To fix this problem, we can combine the troublesome step sizes in pairs and fill two boxes by using the two step sizes alternately. This requires a little care; combining two step sizes that add up to a multiple of $5$ will still not work. But we can combine step sizes of $0$ with $2$, $4$ with $5$, and $6$ with $8$.
For example, one box with alternating steps of $0$ and $2$ would contain
$$
\big[\ (1,1)\quad (1,3)\quad (3,3)\quad (3,5)\quad (5,5)\quad
(5,7)\quad (7,7)\quad (7,9)\quad (9,9)\quad (9,1)\ \big],
$$
and the other would contain
$$
\big[\ (2,2)\quad (2,4)\quad (4,4)\quad (4,6)\quad (6,6)\quad
(6,8)\quad (8,8)\quad (8,10)\quad (10,10)\quad (10,2)\ \big].
$$
This way all ten boxes are filled: one each with steps of $1$, $3$, $7$ and $9$ and two each with pairs of $0$ with $2$, $4$ with $5$, and $6$ with $8$.
EDIT: As Jaap Scherphuis points out, the combination of pairs with steps $4$ and $5$ does not work: after ten pairs are placed in a box, the total of steps is $45$, not a multiple of ten.
To fix this, the revised plan is to combine the pairs with steps $3$, $4$ and $5$ into three boxes, with the successive step sizes in each box as follows:
$$
\text{First Box: }3, 4, 5, 3, 4, 5, 3, 4, 5, 4\\
\text{Second Box: }3, 4, 5, 3, 4, 5, 3, 4, 4, 5\quad\\
\text{Third Box: }3, 4, 5, 3, 4, 5, 5, 3, 5, 3\
$$
It is a bit tricky to make sure that the steps within each box don't cause a pair to be repeated and that the three boxes together can be filled with non-overlapping sets of pairs; the starting points matter as well as the steps.
To make sure this works, here are the contents of the boxes:
$$
\big[\ (7,10)\quad (10,4)\quad (4,9)\quad (9,2)\quad (2,6)\quad
(6,1)\quad (1,4)\quad (4,8)\quad (8,3)\quad (3,7)\ \big]
$$
$$
\big[\ (4,7)\quad (7,1)\quad (1,6)\quad (6,9)\quad (9,3)\quad
(3,8)\quad (8,1)\quad (1,5)\quad (5,9)\quad (9,4)\ \big]
$$
$$
\big[\ (3,6)\quad (6,10)\quad (10,5)\quad (5,8)\quad (8,2)\quad
(2,7)\quad (7,2)\quad (2,5)\quad (5,10)\quad (10,3)\ \big]
$$
I think that does it.