271. Find a four-digit number $\overline{a b c d}$, divisible by 11, under the condition: $b+c=a$ and $\overline{b c}$ is a perfect square.
271. Since $\overline{b c}$ is a perfect square, then $c=0,1,4,5,6,9$.

Given that $\overline{a b c d}$ is divisible by 11, then $a+c-b-d$ can take the values 0 and $\pm 11$. Let's consider these cases:

1) $\left\{\begin{array}{l}a+c=b+d \\ b+c=a\end{array}\right.$, from this $2 c=d$, therefore, $c=0$, $d=0 ; c=1, d=2 ; c=4, d=8$.

Since $\overline{b c}$ is a perfect square, then for $c=0$ we get $b=0$, but then $a=0$; this case is excluded. For $c=1$ and $d=2$ we get $b=8$ and 0; then $a=9$ and $a=1$. We get the numbers 9812 and 1012. For $c=4$ and $d=8$: 1) $b=0$, $a=4$ and 2) $b=6, a=10$ (the second case is excluded). The desired number is 4048.
2) $\left\{\begin{array}{l}a+c-b-d=11 \\ b+c=a\end{array}\right.$, from this $2 c-d=11$, therefore, $c>5, d$ - odd: $c=6, d=1 ; c=9, d=7$.

For $c=6$ and $d=1$ we get $b=3, a=9$. The desired number is 9361.

For $c=9$ and $d=7$ we get $b=0$ and 4, if $b=0$, then $a=9$, the desired number is 9097; if $b=4$, then $a=13$, this case is excluded.
3) $\left\{\begin{array}{l}b+d-a-c=11 \\ b+c=a\end{array}\right.$, from this $d-2 c=11$, which is impossible, since $d \leqslant 9$.

Answer. 9812, 1012, 4048, 9361, 9097.