Possible?

$\large 1 \; \square \; \dfrac{1}{2} \; \square \; \dfrac{1}{3}\; \square \; \dfrac{1}{4}\; \square \; \dfrac{1}{5} \; \square \; \dfrac{1}{6} \; \square \; \dfrac{1}{7} \; \square \; \dfrac{1}{8} \; \square \; \dfrac{1}{9} \; \square \; \dfrac{1}{10} \; \square \; \dfrac{1}{11} \; \square \; \dfrac{1}{12}= 0$

Can you fill the boxes with $+ \text{ and } -$ to make this equation true?