We are creating an infinite table of integers with the following procedure: The first column contains the terms of an arithmetic sequence in ascending order, whose first term is 4 and the common difference is 3. For every $k \geq 1$, the $k$-th row is the arithmetic sequence whose common difference is $2k+1$. Where can the 1994 and the $1995$ be found in this table?
Let the element in the $m$-th row and $n$-th column of the table be denoted by $a_{m, n} (m, n \geq 1)$. The first column forms an arithmetic sequence with the first term 4 and a common difference of 3, thus

$$
a_{m, 1}=4+3 \cdot(m-1)=1+3 m
$$

The $m$-th row is also an arithmetic sequence with the first term $a_{m, 1}$ and a common difference of $2 m+1$, that is,

$$
a_{m, n}=a_{m, 1}+(2 m+1)(n-1)=1+3 m+2 m n-2 m+n-1=2 m n+m+n
$$

Suppose $a_{m, n}=k$, then

$$
k=2 m n+m+n, 2 k+1=4 m n+2 m+2 n+1=(2 m+1)(2 n+1)
$$

For the case $k=1994$, this gives

$$
3989=(2 m+1)(2 n+1)
$$

However, 3989 is a prime number and cannot be factored into the product of two integers greater than 1. Therefore, 1994 does not appear in the table.

For the case $k=1995$,

$$
3991=(2 m+1)(2 n+1)
$$

Since the prime factorization of 3991 is $13 \cdot 307$, the expression (1) can be written in two ways: $2 m+1=13$ and $2 n+1=307$, which gives $m=6$ and $n=153$, and vice versa, $2 m+1=307$ and $2 n+1=13$, which gives $m=153$ and $n=6$. Therefore, 1995 appears in two places: as the 153rd element in the 6th row and as the 6th element in the 153rd row.

Based on the work of Zoltán Radnai (Budapest, Alternatív Közgazd. Gimn., II. o.t.)