Which sum of sum of arithmetic progressions?

Nice question!

It is interesting to note the following:

If we define hypersums $\displaystyle R_q = \sum_{i=1}^q T_i$ and $\displaystyle Z_k = \sum_{i=1}^k R_i$ , then

$\frac {a_1+a_p} 2= \;\displaystyle \frac {S_p} {\dbinom p 1} = \frac {T_m} {\dbinom {m+1}2} = \frac {R_q} {\dbinom {q+2}3} = \frac {Z_k} {\dbinom {k+3}4}$

where $\dfrac {p-1}2 = \dfrac {m-1}3= \dfrac {q-1}4 = \dfrac {k-1}5$ .

It can be shown this relationship is true for higher order hypersums as well.

For this case, solving $\dfrac {p-1}2 = \dfrac {m-1}3$ gives $m=1+\dfrac 3 2 (p-1)$ and substituting $p = 2019$ gives $m = \color{#D61F06}{3028}$ .

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Details added:

Note that

$\small\begin{aligned} S_n &=\sum_{r=1}^n a_r &&=\binom n1 a \qquad+ \binom n2 d &&= \binom n1\qquad \left(a + \frac {n-1}2 \cdot d\right)\\ T_m &=\sum_{n=1}^m S_n &&=\binom {m+1}2 a + \binom {m+1}3 d &&= \binom {m+1}2 \left(a+ \frac {m-1}3\cdot d\right)\\ R_q &=\sum_{m=1}^q T_m &&= \binom {q+2}3 a+\binom {q+2}4 d &&=\binom {q+2}3 \left( a +\frac {q-1}4 \cdot d\right)\\ Z_k &=\sum_{q=1}^k R_q &&= \binom {k+3}4 a+\binom {k+3}5 d &&=\binom {k+3}4 \left( a +\frac {k-1}5 \cdot d\right)\\ \end{aligned}$

Let the first term of the progression be a and the common difference be d. Then $S_{2019}$ =2019a+2037171d; $T_{m}$ = $\dfrac{1}{3}$ m(m+1)[3a+(m-1)d]. Comparing the coefficients of a and d in the two equations, we get $\dfrac{2019}{3}$ = $\dfrac{2037171}{m-1}$ or m=1+3027=3028.