How to find value of $x+y+z+u+v+w$ let $x,y,z,u,v,w$ be positive integer numbers,and such
$$1949(xyzuvw+xyzu+xyzw+xyvw+xuvw+zuvw+xy+xu+xw+zu+zw+vw+1)=2004(yzvw+yzu+yzw+uvw+y+u+w)$$
Find this value of
$$x+y+z+u+v+w=?$$
My try:
maybe use

$$(x+1)(y+1)(z+1)(u+1)(v+1)(w+1)=(xyz+xy+yz+xz+x+y+z+1)(uvw+uv+uw+vw+u+v+w+1)$$


This is not a solution but a little progress towards the solution pointed out in Ewan Delanoy's comment to show that $x=u=1$ and $y \ge 36$.
Let each side of original equation be $2004 \times 1949 \times k$. Multiplying the parentheses on right-hand-side with $x$ and subtracting from the parentheses on left-hand-side gives $xyzvw(u−1)+xyvw+zuvw+zu+zw+vw+1=(2004−1949x)k$ . The left-hand-side is positive so right-hand-side must be positive so $x=1$.
If $x=1$,
$yzuvw+yzu+yzw+yvw+uvw+zuvw+y+u+w+zu+zw+vw+1=2004k$,
$yzvw+yzu+yzw+uvw+y+u+w=1949k$.
Since $z\ge 1$, first eqn yields $y(zuvw+zu+zw+vw)+2(uvw+u+w)+y+vw+1 \le 2004k$ and dividing by second eqn yields $\frac{y(zuvw+zu+zw+vw)+2(uvw+u+w)+y+vw+1}{y(zvw+zu+zw+1)+uvw+u+w} \le \frac{2004}{1949}$ which implies $\frac{zuvw+zu+zw+vw}{zvw+zu+zw+1} < \frac{2004}{1949}$ or $1949uzvw+1949vw<2004zvw+55(zu+zw)+2004$.
Substituting $zu \le uzvw$, $zw \le zvw$, and $1 \le vw$, $1949uzvw+1949vw<2059zvw+55uzvw+2004vw$ or $1894uz<2059z+55$ which precludes $u \ge 2$.
With $x=u=1$, the two equations reduce to
$yzvw+yz+yzw+yvw+vw+zvw+y+w+z+zw+vw+2=2004k$,
$yzvw+yz+yzw+vw+y+w+1=1949k$
and the difference of the two equations is
$yvw+zvw+z+zw+vw+1=55k$.
The ratio of the last two equations is
$\frac{(yz+1)(vw+w+1)+y}{(z+1)(vw+w+1)+yvw-w} = \frac{1949}{55}$.
But $y \le yvw - w$ unless $v=w=1$. Even if $v=w=1$, $y/(yvw - w) \le 1959/55$ so 
$\frac{yz+1}{z+1} \ge \frac{1949}{55}$ so $y \ge 36$.
