Series of natural numbers which has all same digits
Solution 1:
I found that it is known that the only triangular numbers (i.e. sums $1+2+\cdots+n$) which are also "repdigits" (i.e. all the same digit in base 10) are, as listed in the O.E.I.S as sequence A045914 those in the list $$0,\ 1,\ 3,\ 6,\ 55,\ 66,\ 666.$$ As you see they are including $0$ (the "empty sum" of digits, or the $0^{th}$ triangular number.) They also include the single digit numbers $1,3,6$ since technically they are repdigits of only one digit, and are also triangular. Thus it is known that the three you mention are the only ones of length 2 or more.
In a related article/list A213516 of triangular numbers with at most two different digits, in the discussion they refer definitely to the fact that the above list is complete for the one digit triangular numbers. Also at the first reference A045914 there is a citation of a paper by D. Ballew and R. Wagner which appeared in J. Rec. Math. Vol 8 (2) p 96, year 1975-76, titled "Repdigit Triangular Numbers". I don't have access to that, but perhaps in that paper they have a proof that the above list is complete.
Solution 2:
This is a partial answer. I can't access to the article of Ballew and Weger, so I'm trying to prove this myself.
Let's say that $9n(n+1)=2dK_r$ where $K_r=10^r-1$ for $r\ge 2$.
Some partial results:
The digit $d$ is $1$, $5$ or $6$. Indeed, we know that $9+8dK_r$ must be a perfect square, so putting $d\in\{2,3,4,7,8,9\}$ and taking mod ${100}$ we get that $$9+8dK_r\equiv9-8d\in\{1,93,85,77,69,61,53,45,37\}\pmod{100}$$ There are only three quadratic residues mod $100$ in this list, namely $1$, $69$ and $61$, corresponding to $d=1,5,6$.
If $d=5$ then $r$ is even. Because if $r=2t+1$ for some $t\ge 1$, we get that $$9+40(10^{2t+1}-1)=4\cdot100^{t+1}-31$$ should be a perfect square, but since $$(2\cdot10^{t+1}-1)^2<4\cdot100^{t+1}-31<(2\cdot10^{t+1})^2$$ it can not.
Some dead ends
Taking mod $5^r$ for greater values of $r$ seems to be useless, because $9-8d=1,-31,-39$ are quadratic residues mod $5^r$ for every $r$.
The suggestion of @Peter about comparing the fractional parts of $A=\sqrt{9n(9n+1)}$ and $B=\sqrt{2dK_r}$, doesn't seem to lead anywhere. The fractional part of $B$ is difficult to control. Nevertheless, I have computed some values for $d=1,6$ and the distribution of the fractional parts of $\sqrt{2dK_r}$ does not seem entirely random.