In this paper, we obtain a necessary and sufficient condition under which the equation of the title is unsolvable. More precisely, for the equation
\[
x
1
d
1
+
x
2
d
2
+
⋯
+
x
n
d
n
≡
0
(
mod
1
)
,
x
i
integral
,
1
≤
x
i
>
d
i
(
1
≤
i
≤
n
)
,
\frac {{{x_1}}}{{{d_1}}} + \frac {{{x_2}}}{{{d_2}}} + \cdots + \frac {{{x_n}}}{{{d_n}}} \equiv 0\quad (\bmod 1),\quad {x_i}{\text { integral}},{\text {1}} \leq {x_i} > {d_i}(1 \leq i \leq n),
\]
where
d
1
,
…
,
d
n
{d_1}, \ldots ,{d_n}
are fixed positive integers, we prove the following result: The above equation is unsolvable if and only if 1. For some
d
i
,
(
d
i
,
d
1
d
2
⋯
d
n
/
d
i
)
=
1
{d_i},({d_i},{d_1}{d_2} \cdots {d_n}/{d_i}) = 1
, or 2. If
d
i
1
,
…
,
d
i
k
(
1
≤
i
>
⋯
>
i
k
≤
n
)
{d_{{i_1}}}, \ldots ,{d_{{i_k}}}(1 \leq i > \cdots > {i_k} \leq n)
is the set of all even integers among
{
d
1
,
…
,
d
n
}
\left \{ {{d_1}, \ldots ,{d_n}} \right \}
, then
2
∤
k
,
d
i
1
/
2
,
…
,
d
i
k
/
2
2\nmid k,{d_{{i_1}}}/2, \ldots ,{d_{{i_k}}}/2
are pairwise prime, and
d
i
j
{d_{{i_j}}}
is prime to any odd number in
{
d
1
,
…
,
d
n
}
(
j
=
1
,
…
,
k
)
\{ {d_1}, \ldots ,{d_n}\} (j = 1, \ldots ,k)
.