6. In the sequence of natural numbers starting from 1, certain numbers are colored red according to the following rules. First, color 1; then color two even numbers 2, 4; then color the three consecutive odd numbers closest to 4, which are $5, 7, 9$; then color the four consecutive even numbers closest to 9, which are $10, 12, 14, 16$; then color the five consecutive odd numbers closest to 16, which are $17, 19, 21, 23, 25$. Continue this pattern, resulting in a red subsequence $1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, \cdots$ - The 1996th number in the red subsequence, starting from 1, is $\qquad$
6. 3929 .

The first time, color one number red: $1,1=1^{2}$;
The second time, color 2 numbers red: $2,4,4=2^{2}$;
The third time, color 3 numbers red: $5,7,9,9=3^{2}$;
Guessing, the last number colored red in the $k$-th time is $k^{2}$. Then the $k+1$ numbers colored red in the $(k+1)$-th time are:
$$
k^{2}+1, k^{2}+3, k^{2}+5, \cdots, k^{2}+2 k-1, k^{2}+2 k+1 \text {. }
$$

The last number is $(k+1)^{2}$. According to this rule, the first $k$ segments have colored a total of
$$
\begin{array}{l}
1+2+3+\cdots+(k+1)+k=\frac{k(k+1)}{2} \text { numbers. } \\
\text { Solve } \frac{k(k+1)}{2} \leqslant 1996, \text { we get } k \leqslant 62 .
\end{array}
$$

At this point, the first 62 segments have colored a total of $\frac{62 \times 63}{2}=1953$ numbers. The last number in the 62nd segment is $62^{2}=3844$, which is the 1953rd red number. The 1954th red number is 3845. And $1996-1953=$ 43, so the 1996th red number is