If $\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}$, then $N=$
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992$
Note that for all integers $n\neq 0$, 
\[\frac{(n-1)+n+(n+1)}{n}=3.\]
Thus, we must have $N=1991\rightarrow \boxed{\text{D}}$.