6. For $n \in \mathbf{N}_{+}$, if $2^{n} n+1$ is a multiple of 3, then the set of remainders when $n$ is divided by 6 is
6. $\{1,2\}$.
$$
\begin{array}{l}
2^{n} n+1=(3-1)^{n} n+1 \\
=3 M+(-1)^{n} n+1(M \in \mathbf{N}) .
\end{array}
$$

If $3 \mid\left(2^{n} n+1\right)$, then $3 \mid \left[(-1)^{n} n+1\right]$.
When $n=6 k\left(k \in \mathbf{N}_{+}\right)$,
$$
(-1)^{n} n+1=6 k+1 \equiv 1(\bmod 3) \text {; }
$$

When $n=6 k+1(k \in \mathbf{N})$,
$$
(-1)^{n} n+1=-6 k \equiv 0(\bmod 3) \text {; }
$$

When $n=6 k+2(k \in \mathbf{N})$,
$$
(-1)^{n} n+1=6 k+3 \equiv 0(\bmod 3) \text {; }
$$

When $n=6 k+3(k \in \mathbf{N})$,
$$
(-1)^{n} n+1=-6 k-2 \equiv 1(\bmod 3) \text {; }
$$

When $n=6 k+4(k \in \mathbf{N})$,
$$
(-1)^{n} n+1=6 k+5 \equiv 2(\bmod 3) \text {; }
$$

When $n=6 k+5(k \in \mathbf{N})$,
$$
(-1)^{n} n+1=-6 k-4 \equiv 2(\bmod 3) \text {. }
$$

In summary, $3 \mid\left(2^{n} n+1\right)$ if and only if $n=6 k+1$ or $6 k+2$ $(k \in \mathbf{N})$.