, 2026

We begin by determining the values of the digits $B$ and $C$ when $A<9$. At the end of the solution we
will show that the values of $B$ and
$C$ are the same when $A=9$.

Suppose that when $A<9$, the value
of $0.ABC$ rounded to the nearest
tenth is $0.T00$.

Since $0.T00$ is $0.024$ greater than $0.ABC$, then $0.ABC+0.024=0.T00$.

The thousandths digit of $0.T00$ is
0, and so the sum of the thousandths digits, $C$ and $4$, ends in $0$. Since $C$ is at most $9$, then $C+4=10$, and so $C=6$.

In this case, the 'carry' from the thousandths column to the hundredths
column is $1$.

The hundredths digit of $0.T00$ is
$0$, and so the sum of the hundredths
digits, $B$ and $2$, added to the carry of $1$, ends in $0$. Since $B$ is at most $9$, then $B+2+1=10$, and so $B=7$.

In this case, the carry from the hundredths column to the tenths column
is $1$, and so $T=A+1$.

For any choice of the non-negative digit $A$, where $A<9$, $0.A76+0.024=0.T00$ where $T=A+1$.

As examples, $0.876+0.024=0.900$ and
$0.076+0.024=0.100$.

Finally, we show that when $A=9$, the
values of $B$ and $C$ remain the same.

The value of $0.9BC$ when rounded to
the nearest tenth is greater than $0.9BC$, and therefore is equal to $1$.

In this case, $0.9BC+0.024=1$ or
$0.9BC=1-0.024=0.976$, and so $B=7$ and $C=6$.

Therefore, for all possible values of the non-negative digit $A$, we get $B+C=7+6=13$.