The structure of the tree need not be the one described. And counting recipients by layers is not the most efficient or reliable way.
We have $10000$ people sending letters, each to $5$ "new" people. On the assumption, unfortunately not entirely safe, that any letter sent is received, there are $50000$ recipients.
Note that $9999$ of the people who send letters are letter recipients. The rest of the recipients do not send letters.
For simplicity, assume $n$ is odd and divisible by $3$.
While a necessary and sufficient condition as given in Mike's answer is good to have, simply saying "the construction is too complicated to reproduce in an answer" does not preclude showing the simpler parts of that construction.
Theorems 5 and 6 in Ray-Chaudhuri and Wilson give explicit constructions of Kirkman triple systems on $3q$ and $2q+1$ points where $q=6t+1$ is a prime power. For $3q$ points, label them as $(x,j)$ where $x\in\mathbb F_q$ and $j\in\{0,1,2\}$; furthermore let $g$ be a primitive element of $\mathbb F_q$. Define triples
$$Z=\{(0,0),(0,1),(0,2)\}\\
B_{i,j}=\{(g^i,j),(g^{i+2t},j),(g^{i+4t},j)\},i\in[0,t),j\in\{0,1,2\}\\
A_i=\{(g^i,0),(g^{i+2t},1),(g^{i+4t},2)\},i\in[0,6t)$$
Then one of the parallel classes (time periods) is $Z\cup\{B_{i,j}\}\cup\{A_i:\lfloor i/t\rfloor\text{ odd}\}$ where indices have their full range as defined above unless stated otherwise. $q$ parallel classes are obtained from this class by shifting (adding the same element of $\mathbb F_q$ to the first part of every element in every block). The remaining classes are made of an $A_i$ where $\lfloor i/t\rfloor$ is even and all its shifts.
For $2q+1$ points label them similarly as above, except that $j\in\{0,1\}$ and the last point is called $\infty$; let $m$ solve $2g^m=g^t+1$. One parallel class consists of blocks
$$\{(0,0),(0,1),\infty\}\\
\{(g^{i+2kt},0),(g^{i+2kt+t},0),(g^{i+2kt+m},1)\},i\in[0,t),k\in\{0,1,2\}\\
\{(g^{i+m+t},1),(g^{i+m+3t},1),(g^{i+m+5t},1)\},i\in[0,t)$$
and all the rest are obtained by shifting, except that $\infty$ does not change.
Here is an illustration of these blocks and classes when $t=2$, so $q=13$, $g=2$ and $m=8$:
A Kirkman triple system on $9$ points is given by the parallel lines of the Hesse configuration:
$$(012,345,678)\\
(036,147,258)\\
(048,156,237)\\
(057,138,246)$$
Best Answer
W1hat you are looking for is equivalent to a row-complete Latin square. A Latin square is an $n\times n$ table filled with the numbers $1$ to $n$ so that each row and column has no repeats. A Latin square is further called row-complete if the $n\times (n-1)$ ordered pairs of numbers which occur horizontally adjacently in the square are also all distinct.
To apply this to your problem, given a row-complete Latin square, numbered the papers from $1$ to $n$. The rows represent people, the columns represent time, and the label at any row and column represents the paper that that person signs at the time. The Latin square conditions imply everyone has a different paper, and that everyone signs all papers. The row-complete condition ensures everyone receives their papers from a different person each time.
It is easy to solve the problem when $n$ is even. First, everyone passes their paper one spot to the left, then two spots to the right, then three spots to the left, then four spots to the right, and so on, ending with passing $(n-1)$ spots to the left.
It is proven in [Higgam] that when $n$ is odd and composite, a row complete Latin square of order $n$ exists. The construction uses finite fields. The construction is provided in that paper, and is also available in this survey by Otis, available online at https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS10, in section 3.6.
All that remains are odd primes. No row-complete Latin square has been found for any odd prime order. Furthermore, exhaustive searches have shown there are no squares of orders $3,5$ and $7$. It is an open problem for higher odd primes.
In the odd prime case, where no row-complete Latin square is known to exist, the following method ensures that there is at most one repeat among the people who pass you the papers. Let $n=2m+1$. The algorithm differs slightly when $m$ is even vs odd.
In either case, the first $m$ steps are the same: start with passing one left, two right, three left, four right, and so on, ending at passing $m$ in a direction depending on the parity of $m$.
Next, if $m$ is odd...
If instead $m$ is even...
Examples
Approximate solution for $n=7$: (note the each explanation is for each column despite being written next to the row).
Approximate solution for $n=9$:
Exact solution for $n=9$: