2. The sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with all terms being positive integers, and the common difference $d \in \mathbf{N}_{+}$. If the sum of any two terms in $\left\{a_{n}\right\}$ is still a term in the sequence, and $a_{1}=6$, then $d=$ $\qquad$
$$
2.1,2,3,6 \text {. }
$$

For any two terms $a_{m} 、 a_{k}$, their sum $a_{m}+a_{k}=a_{n}$, i.e.,
$$
2 a_{1}+(m+k-2) d=a_{1}+(n-1) d \text {. }
$$

Rearranging gives $a_{1}=(n-m-k+1) d=6$.
Since $d \in \mathbf{N}_{+}, n-m-k+1$ is an integer, thus, $d=1,2,3,6$.