2. Let $a_{1}, a_{2}, \cdots, a_{k}$ be $k$ distinct positive integers, and $a_{1}+a_{\varepsilon}+\cdots+a_{k}=1995$. Then, the maximum value of $k$ is $\qquad$
2. $\because a_{1}, a_{2}, \cdots, a_{4}$ are all distinct, without loss of generality, let $a_{1}<a_{2}<\cdots<a_{k}$. Since $a_{1}, a_{2}, \cdots, a_{k}$ are all positive integers, we have $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{k} \geqslant k$, thus $1+2+\cdots+k \leqslant a_{1}+a_{2}+\cdots+a_{k}=$ 1995, which means $\frac{k(k+1)}{2} \leqslant 1995$, solving this yields $1 \leqslant k \leqslant 62$. On the other hand, when $a_{1}=1, a_{2}=2, \cdots, a_{61}=61, a_{62}=104$, such $a_{1}$, $a_{2}, \cdots, a_{4}$ indeed satisfy the conditions of the problem, therefore the maximum value of $k$ is 62.