Reciprocal sum of triangular numbers is 1. n∈N
Triangular numbers set is $$\{1,3,6,10,15,21,28,36,45,55,\dots,\frac{n(n+1)}2,\dots\}$$
n≥3
I solved the equation   $$\frac 1{x_1} +\frac1{x_2} +\cdots+\frac1{x_n} =1$$      for triangular numbers:
$$\begin{align}n&=3,(3,3,3)\\
n&=4,(3,3,6,6)\\
n&=5,(3,6,6,6,6)\\
n&=6,(6,6,6,6,6,6)\\
n&=7,(3,6,10,10,10,10,10)      \text{ and } \\   &\quad(3,3,15,15,15,15,15)\\  
n&=8,(3,6,6,15,15,15,15,15)\\
n&=9,(6,6,6,6,15,15,15,15,15)\\
n&=10,(10,10,10,10,10,10,10,10,10,10)\end{align}$$
Question is,
how can I find the solution of the equation for triangular numbers for other $n$?
 A: Not an answer, just expanding on my ideas in comments.
Let $x_i=\frac{m_i(m_i+1)}{2}.$ Then $\frac{1}{x_i}=\frac{2}{m_i(m_i+1)}=\frac2{m_i}-\frac2{m_i+1}.$ Then your equation can be written as: $$\sum_{i=1}^{n} \frac{1}{m_i}=\frac 12+\sum_i \frac1{m_i+1}$$
If the largest $m_i$ has $m_i+1=p^k>2$ for some prime $p,$ then the number of $i$ with $m_i=p^k-1$ must be a multiple of $p.$ This is because the right side will still have $p^k$ as a factor in the denominator if not, and it can't be a factor on the left side.
If the largest $m_i$ has $m_i=p^k>2$ then the number of $m_i=p^k$ must be congruent to the number of $m_i=p^k-1$ modulo $p.$
So when the largest $x_i=10,m_i=4$ then the number of $10$s must be congruent to the number of $6$s modulo $2,$ and the number of $10$s must be a multiple of $5.$
If $m_i=5,x_i=15$ is the largest, the number of $15$s must be congruent to the number of $10$s modulo $5.$ We have $a$ $10s$ and $5b+a$ $15?$s, you get the sum of those is:
$$a\left(\frac1{10}+\frac1{15}\right)+\frac{5b}{15}=\frac a6+\frac b3.$$
When $a+2b\leq 6,$ you get solutions: $((6-a-2b)\times 6,a\times 10,(a+5b)\times 15).$ This gives an answer for $n=6-a-2b+a+a+5b=6+a+3b.$
When $6-a-2b>2,$ we can replace any pairs of sixes with a $3.$
So when $a=2,b=1,$ we get another solution for $n=10,$ $x_*(3,2\times 10,7\times 15)$ or $m_*=(2,2\times 4,7\times 5).$
When the largest $m_i=6, x_i=21$ then there must be a multiple of $7$ of them. Then you get $\frac{7b}{21}=\frac b3.$ You can $\frac13$ and $\frac23$ in quite a few ways with the smaller $m_i.$
$m_i=7$ is interesting because $m_i$ and $m_i+1$ are prime powers, so if the largest $x_i=28,$ then the number of $28$s must be even and congruent to the number of $21$s modulo $7.$ If there are $a$ $21$s] and $a+7b$ $28$s, then $a+b$ must be even, the partial sum $\frac{a}{12}+\frac{b}{4}=\frac{a+3b}{12}=\frac{b+\frac{a+b}2}6$ and thus we can make thirds and sixths out of the smaller $m_i.$
But this approach will really only work for small $m_i.$ Prime powers became rarer when $m$ increases. When $m_i=14,$ neither $m_i$ nor $m_i+1$ is a prime power. You might still be able to do something with that, but it gets more complicated.

Solutions for all $n>2:$
If $T(k)=\frac{k(k+1)}2,$ we have: $$\sum_{k=2}^{m-1}\frac1{T(k)}=1-\frac{2}{m},$$ we can add $m+1$ copies of $\frac1{T(m)}$ to get a total of $1,$ yielding a solution $(T(2),T(3),\dots,T(m-1),(m+1)\times T(m))$ for $n=2m-1.$ That gives a solution for all odd $n.$
For $n=11,$ this becomes $(3,6,10,15,7\times 21).$
We can get a solution for $n=2m$ by first solving for $2m-1,$ then replacing the $3$ with two $6$s.
So we can find a solution for all $n>2.$
Since $\sum_{m=2}^{\infty}\frac{1}{T(m)}=1,$ there is no $n>2$ for which there no repeating values. For $n$ odd, we see there is a solution with only one repeated value. For $n=2m$ even, our solution has potentially two repeated. values, $T(3)$ and $T(m).$
There are cases when $n=T(k)$ is even where we have a solution with only one value, such as $(6,6,6,6,6,6)$ and $(10\times 10).$ But are those the only even cases where all but one value occurs only once?
