Floors and Ceilings 1- Floor and Ceiling Together

$Case\quad 1-\quad If\quad x\quad is\quad an\quad integer,\\ then\quad \left\{ x \right\} =0\quad and\quad thus\quad at\quad least\quad one\quad of\quad \left\lfloor x \right\rfloor \quad or\quad \left\lceil x \right\rceil \quad must\quad \\ be\quad zero.\\ But\quad as\quad x\quad is\quad an\quad integer,\quad \left\lfloor x \right\rfloor =\left\lceil x \right\rceil =0\\ Hence\quad only\quad possible\quad integervalue\quad of\quad x=\left\lfloor x \right\rfloor +\left\{ x \right\} =0\\ \\ \\ \\ Case\quad 2-\quad If\quad x\quad is\quad not\quad an\quad integer,\\ then\quad \left\lceil x \right\rceil =\left\lfloor x \right\rfloor +1.\quad Let\quad \left\lfloor x \right\rfloor =t,\quad so\quad we\quad have\quad \left\{ x \right\} =\frac { t\left( t+1 \right) }{ 2015 } \\ Now\quad since\quad 0<\left\{ x \right\} <1\quad for\quad non-integer\quad values\quad of\quad x,\quad we\quad \\ get\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0<\frac { t\left( t+1 \right) }{ 2015 } <1\\ We\quad need\quad not\quad worry\quad about\quad LHS\quad of\quad the\quad inequation\quad as\quad it\quad is\quad \\ always\\ greater\quad than\quad zero\quad for\quad non\quad zero\quad integer\quad values\quad of\quad t.\\ The\quad RHS\quad yields\quad { t }^{ 2 }+t-2015<0\quad or\quad { \left( t+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 8061 }{ 4 } <0,\quad or\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad -\frac { \sqrt { 8061 } }{ 2 } -\frac { 1 }{ 2 } \quad <\quad t\quad <\quad \frac { \sqrt { 8061 } }{ 2 } -\frac { 1 }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad or\quad approximately\quad -45.39\quad <\quad t\quad <\quad 44.39\\ Hence\quad integer\quad values\quad of\quad t\quad are\quad -45,-44,...,-2\quad and\quad \\ 1,2,...,44,\quad yielding\quad a\quad total\quad of\\ 88\quad non-integer\quad values\quad of\quad x.\\ Note\quad that\quad t=\left\lfloor x \right\rfloor \quad cannot\quad be\quad -1\quad \\ as\quad that\quad would\quad yield\quad t+1=\left\lceil x \right\rceil =0,\\ ultimately\quad yielding\quad \left\{ x \right\} =0,\quad implying\quad that\quad x\quad is\quad an\quad integer\quad \\ whereas\quad only\quad permitted\\ integer\quad value\quad of\quad x\quad is\quad 0.\\ Hence\quad x=\left\lfloor x \right\rfloor +\left\{ x \right\} \quad =\quad t+\frac { t\left( t+1 \right) }{ 2015 } =\frac { { t }^{ 2 }+2016t }{ 2015 } \\ Summing\quad this\quad over\quad from\quad t=-45\quad to\quad t=-2\quad and\quad again\quad from\quad \\ t=1\quad to\quad t=44,\\ we\quad get\quad S=-\frac { 2053150 }{ 2015 } +\frac { 2025210 }{ 2015 } =-\frac { -27940 }{ 2015 } \cong -13.866004...\\ Hence\quad n=88+1=89\quad and\quad \left\lfloor 1000S \right\rfloor =-13867\quad and\quad our\quad answer\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 89-13867=\quad \boxed { -13778 } \\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$