The answer is yes. We construct an infinite sequence $1=n_{1}<2=n_{2}<n_{3}<\ldots$ so that $n_{i}+n_{j}$ is square-free for all $i<j$. Suppose that we already have some numbers $n_{1}<\ldots<n_{k}(k \geq 2)$, which satisfy this condition and find a suitable number $n_{k+1}$ to be the next element of the sequence. We will choose $n_{k+1}$ of the form $n_{k+1}=1+M x$, with $M=\left(\left(n_{1}+\ldots+n_{k}+2 k\right)!\right)^{2}$ and some positive integer $x$. For $i=1,2, \ldots, k$ we have $n_{i}+n_{k+1}=1+M x+n_{i}=\left(1+n_{i}\right) m_{i}$, where $m_{i}$ and $M$ are co-prime, so any perfect square dividing $1+M x+n_{i}$ is co-prime with $M$. In order to find a suitable $x$, take a large $N$ and consider the values $x=1,2, \ldots, N$. If a value $1 \leq x \leq N$ is not suitable, this means that there is an index $1 \leq i \leq k$ and some prime $p$ such that $p^{2} \mid 1+M x+n_{i}$. For $p \leq 2 k$ this is impossible because $p \mid M$. Moreover, we also have $p^{2} \leq 1+M x+n_{i}<$ $M(N+1)$, so $2 k<p<\sqrt{M(N+1)}$. For any fixed $i$ and $p$, the values for $x$ for which $p^{2} \mid 1+M x+n_{i}$ form an arithmetic progression with difference $p^{2}$. Therefore, there are at most $\frac{N}{p^{2}}+1$ such values. In total, the number of unsuitable values $x$ is less than $$\begin{aligned} & \sum_{i=1}^{k} \sum_{2 k<p<\sqrt{M(N+1)}}\left(\frac{N}{p^{2}}+1\right)<k \cdot\left(N \sum_{p>2 k} \frac{1}{p^{2}}+\sum_{p<\sqrt{M(N+1)}} 1\right)< \\ & <k N \sum_{p>2 k}\left(\frac{1}{p-1}-\frac{1}{p}\right)+k \sqrt{M(N+1)}<\frac{N}{2}+k \sqrt{M(N+1)} \end{aligned}$$ If $N$ is big enough then this is less than $N$, and there exist a suitable choice for $x$.