Label the jars $1,2, \ldots, 2017$. I claim that the answer is 2014. To show this, we need both a construction and an upper bound. For the construction, for $1 \leq i \leq 201$, put a coin in the jars $10 i+1,10 i+2, \ldots, 10 i+10$. After this, each of the jars $1,2, \ldots, 2010$ has exactly one coin. Now, put a coin in each of the jars $2008,2009, \ldots, 2017$. Now, the jars $1,2, \ldots, 2007,2011,2012, \ldots, 2017$ all have exactly one coin. This gives a construction for $N=2014$ (where $d=1$). Now, we show that this is optimal. Let $c_{1}, c_{2}, \ldots, c_{2017}$ denote the number of coins in each of the jars. For $1 \leq j \leq 10$, define $$s_{j}=c_{j}+c_{j+10}+c_{j+20}+\ldots$$ Note that throughout the process, $s_{1}=s_{2}=\cdots=s_{j}$. It is also easy to check that the sums $s_{1}, s_{2}, \ldots, s_{7}$ each involve 202 jars, while the sums $s_{8}, s_{9}, s_{10}$ each involve 201 jars. Call a jar good if it has exactly $d$ coins. If there are at least 2015 good jars, then one can check that it is forced that at least one of $s_{1}, s_{2}, \ldots, s_{7}$ only involves good jars, and similarly, at least one of $s_{8}, s_{9}, s_{10}$ only involves good jars. But this would mean that $202 d=201 d$ as all $s_{i}$ are equal, contradiction.