102. Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98.
102. Let $N$ be the desired number. According to the problem, we have:

$$
N=131 k+112=132 l+98
$$

where $k$ and $l$ are positive integers. Since $N$ is a four-digit number, it is clear that:

$$
l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75
$$

Further, we have:

$$
131 k+112=132 l+98 ; \quad 131(k-l)=l-14
$$

From this, it is evident that if $k-l$ is not zero, then $l-14$ in absolute value exceeds 130, which is impossible if $l \leqslant 75$. Therefore, it must be that $k-l=0, k=l$, from which we immediately obtain:

$$
\begin{gathered}
l-14=0, k=l=14 \\
N=131 \cdot 14+112=(132 \cdot 14+98)=1946
\end{gathered}
$$