= Solution
Use positive real object and image distances $u,v$ for a <thin lens>. The required absolute <magnification> is $M=1000/15=200/3$, and the specified LCD-to-screen separation gives $u+v=2000\,\mathrm{mm}$, not $v=2000\,\mathrm{mm}$. Since $v/u=M$,
$$
u=\frac{2000}{1+200/3}=29.5567\,\mathrm{mm},\qquad v=1970.4433\,\mathrm{mm}.
$$
The <thin lens> equation yields
$$
\boxed{f=\frac{uv}{u+v}=29.1198\,\mathrm{mm}.}
$$
Place the projection <lens> $29.56\,\mathrm{mm}$ beyond the LCD towards the screen. Its real image is inverted; the signed <magnification> is $-200/3$.
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