$m,x,y,z$

$0{\implies}~{\boxed{0}}$

$0+1{\implies}~{\boxed{1}}$

$1+2+3{\implies}~{\boxed{6}}$

$3+4+5+6{\implies}~{\boxed{18}}$

$6+7+8+9+10{\implies}~{\boxed{40}}$

$10+11+12+13+14+15{\implies}~{\boxed{75}}$

$15+16+17+18+19+20+21{\implies}~{\boxed{126}}~={\color{#3D99F6}{m}}$

$21+22+23+24+25+26+27+28{\implies}~{\boxed{196}}~={\color{#D61F06}{x}}$

$28+29+30+31+32+33+34+35+36{\implies}~{\boxed{288}}~={\color{#20A900}{y}}$

$36+37+38+39+40+41+42+43+44+45{\implies}~{\boxed{405}}~={\color{#624F41}{z}}$

$45+46+47+48+49+50+51+52+53+54+55{\implies}~{\boxed{505}}$

$\color{#3D99F6}{m}{\color{#333333}{+}}{\color{#D61F06}{x}}{\color{#333333}{+}}{\color{#20A900}{y}}{\color{#333333}{+}}{\color{#624F41}{z}}{\implies}{\color{#333333}{\boxed{1015}}}$

Hence the answer is $\color{#3D99F6}{\boxed{1015}}$ .

Let the $n$ th term of the sequence be $a_n$ . Consider the following:

$\begin{array} {ccrrrrr} n & : & 1 & 2 & 3 & 4 & 5 \\ a_n & : & 1 & 6 & 18 & 40 & 75 \\ \dfrac {a_n}n & : & 1 & 3 & 6 & 10 & 15 \end{array}$

We note that $\dfrac {a_n}n = T_n$ , where $T_n = \dfrac {n(n+1)}2$ is the triangular number. Then $a_n = nT_n = \dfrac {n^2(n+1)}2$ . Let us check if the formula is correct. $a_{10} = \dfrac {10^2(10+1)}2 = 550.$ The formula is correct. Then we have:

$\begin{aligned} m = a_6 & = \frac {6^2(6+1)}2 = 126 \\ x = a_7 & = \frac {7^2(7+1)}2 = 196 \\ y = a_8 & = \frac {8^2(8+1)}2 = 288 \\ z = a_9 & = \frac {9^2(9+1)}2 = 405 \end{aligned}$

Therefore, $m+x+y+z = 126+196+288+405 = \boxed{1015}$ .